Solutions Exercises Lecture 3 – AES1030
1. One mol of a gas expands reversible en isobar from an initial volume,
V1 = 0,22 L, to its final volume, V2 = 20,02 L. Initially the gas has a
temperature of 300 K. For the given process calculate: the temperature
of the gas at its final state T2, q, w, ∆U en ∆H for the following cases
a. The gas is an ideal gas
b. The gas is a real gas and can be described by the following
equation of state:
p ⋅V = n ⋅ R ⋅ T + n ⋅ b ⋅ p
Herein the parameters are: p pressure, V volume, n number of
moles, R the gas constant and b a constant specific for the gas
and equal to: b = 0,02 L mol-1.
Further we know that:
R = 8,314 J mol-1 K-1 and CP = 5/2 R.
Solution
Given: 1mol gas
Reversible and isobaric process ⇒ pex = pgas = p and dp=0 thus p = p1 = p2
Initial state: V1 = 0,22 L, T1 = 300 K; p
End state: V2 = 20,02 L; p
Wanted: T2 , q, w, ∆E en ∆H
 ∂H 
 ∂U 
CP − CV = n ⋅ R


 =0=
 ∂V T
 ∂p T
For the description of the process one starts with the first law of thermodynamics:
dU = dq + dw
Herein, first dw is calculated:
dw = − pex ⋅ dV = − p ⋅ dV
reversible
Ad a) gas is ideal ⇒ p ⋅ V = n ⋅ R ⋅ T
V2
V2
V1
V1
w = ∫ dw = ∫ − p ⋅ dV = − p ⋅ ∫ dV
isobaric
w = − p ⋅ (V2 − V1 )
The pressure druk p is not explicitly given. However, for the initial state the volume and
the temperature are given so that the pressure can be computed using the ideal gas law:
p ⋅V = n ⋅ R ⋅ T ⇒ p =
n ⋅ R ⋅T
V
V

n ⋅ R ⋅ T1
⋅ (V2 − V1 ) = −n ⋅ R ⋅ T1 ⋅  2 − 1
V1
 V1 
J
 20.02 L 
w = −1mol ⋅ 8.314
⋅ 300 K ⋅ 
− 1 = −224478 J
mol ⋅ K
 0.22 L

w = − p ⋅ (V2 − V1 ) = −
Next, the change of the internal energy is calculated. Because the process is isobaric it is
advised to first calculate the change in the enthalpy and from there the change of the
internal energy (Note: actually for an ideal gas it does not matter if you first calculate the
change of the internal energy or of the enthalpy. The reason for this is that an ideal gas
does not react on changes of pressure; an ideal gas depends only on the temperature).
For systems not containing of an ideal gas, the calculation of the change of enthalpy is
more practical because the enthalpy is commonly defined in the T,p-space H = H(p,T) so
that the total differential can be written as
 ∂H 
 ∂H 
dH = 
 dp + 
 dT
 ∂T  p
 ∂p T

 ∂H 
gas ⇒ 
 = 0
 ∂p T


isobaric ⇒ dp = 0
 ⇒ dH = CP ⋅ dT

 ∂H 

general ⇒ CP = 

 ∂T  p 
ideal
Assuming the heat capacity is constant allows the computation of the change of the
enthalpy:
T2
T2
T1
T1
dH = CP ⋅ dT ⇒ ∆H = ∫ dH = ∫ CP ⋅ dT = CP ⋅ ∫ dT = CP ⋅ (T2 − T1 )
For the computation the temperature of the initial and of the end state is needed. The
initial temperature is given, the end temperature can be computed with the ideal gas law
because for the end state the volume and the pressure are known:
n ⋅ R ⋅T
p ⋅V = n ⋅ R ⋅ T ⇒ p =
V
n1 ⋅ R ⋅ T1 n2 ⋅ R ⋅ T2
⇒ p1 = p2 ⇒
=
n1 = n2
V1
V2
T1 T2
V
20.02 L
= ⇒ T2 = T1 ⋅ 2 = 300 K ⋅
= 27300 K
V1 V2
V1
0.22 L
So that:
⇒
5
5
J
∆H = CP ⋅ (T2 − T1 ) = n ⋅ CP ⋅ (T2 − T1 ) = 1mol ⋅ ⋅ R ⋅ ( 27300 − 300 ) K = 1mol ⋅ ⋅ 8.314
⋅ 27000 K
2
2
mol ⋅ K
∆H = 561195 J
Further we know that H = U + pV and thus:
dH = dU + d(pV) = dU + p dV + V dp = dq + dw + p dV + V dp
and for a reversible process (pex = p):
dH = dq – p dV + p dV + V dp = dq + V dp
and additionally for an isobaric process:
dH = dq = 561195 J.
Finally, the change of the internal energy is computed using the first law of
thermodynamics:
dU = dq + dw ⇒ ∆U= q + w = 561195 J –224478 J = 336717 J = 336.7 kJ
Ad b) In exercise b the same process as in exercise a is computed. However, now the
system consist of a real gas described by the following equation of state
p ⋅V = n ⋅ R ⋅ T + n ⋅ b ⋅ p
The approach is thus the same but the assumptions made for an ideal gas do not hold
anymore. Again we start with the first law: dU = dq + dw
And compute in a first step dw:
dw = − pex ⋅ dV = − p ⋅ dV
reversible
V2
V2
V1
V1
w = ∫ dw = ∫ − p ⋅ dV = − p ⋅ ∫ dV
isobaric
w = − p ⋅ (V2 − V1 )
The pressure is not explicitily given but can be computed using the given equation of
state:
n ⋅ R ⋅T
p ⋅ V = n ⋅ R ⋅ T + n ⋅ b ⋅ p ⇒ p ⋅V − n ⋅ b ⋅ p = p ⋅ (V − n ⋅ b ) = n ⋅ R ⋅ T ⇒ p =
V − n ⋅b
J
1mol ⋅ 8.314
⋅ 300 K
n ⋅ R ⋅ T1
mol ⋅ K
w = − p ⋅ (V2 − V1 ) = −
⋅ (V2 − V1 ) = −
⋅ ( 20.02 − 0.22 ) L
L
V1 − n ⋅ b
0.22 L − 1mol ⋅ 0.02
mol
w = −246926 J
Next the change of the enthalpy is computed. The approach and derivation of the
equation is the same as for exercise a. With the assuming that the heat capacity constant
and not a function of the temperature we get:
T2
T2
T1
T1
dH = CP ⋅ dT ⇒ ∆H = ∫ dH = ∫ CP ⋅ dT = CP ⋅ ∫ dT = CP ⋅ (T2 − T1 )
The temperature at the end state is now computed using the given equation of state and
the values of p and V for the end state:
n ⋅ R ⋅T
V − n ⋅b
n ⋅ R ⋅ T1 n2 ⋅ R ⋅ T2
⇒ p1 = p2 ⇒ 1
=
V1 − n1 ⋅ b V2 − n2 ⋅ b
p ⋅V = n ⋅ R ⋅ T + n ⋅ b ⋅ p ⇒ p =
n1 = n2
L
T1
T2
V − n⋅b
mol = 300 K ⋅ 20 L
⇒
=
⇒ T2 = T1 ⋅ 2
= 300 K ⋅
L
V1 − n ⋅ b V2 − n ⋅ b
V1 − n ⋅ b
0.02 L
0.22 L − 1mol ⋅ 0.02
mol
⇒ T2 = 30000 K
20.02 L − 1mol ⋅ 0.02
So that the change of the enthalpy can be computed:
5
5
J
∆H = CP ⋅ (T2 − T1 ) = n ⋅ CP ⋅ (T2 − T1 ) = 1mol ⋅ ⋅ R ⋅ ( 30000 − 300 ) K = 1mol ⋅ ⋅ 8.314
⋅ 29700 K
2
2
mol ⋅ K
∆H = 617314.5 J = 617.3kJ
For the computation of the exchanged heat, we can use the same equation as for exercise
a: dH = dq = 617314,5 kJ.
The change of the internal energy is calculated using the first law:
dU = dq + dw
⇒ ∆U= q + w = 617314.5 J – 246926 J = 370388.5 J = 370.4 kJ
2. Show that for an ideal gas undergoing a reversible and adiabatic
process the following equation holds:
T1  p2 
= 
T2  p1 
1− γ
γ
CP
γ
=
with
CV
Note:
- CP and CV have as unit [ J K-1 mol-1]
- Start with the first law of thermodynamics; combine the 1st law
with the description of dU for a reversible adiabatic process and
the ideal gas law.
 a
c
- Remember that: ln  = ln( a ) − ln( b) , c ⋅ ln ( a ) = ln ( a ) and
b
(
exp ln ( a )
C
)=a
c
Solution:
First, we collect all the given data and information:
The system consists of an ideal gas and undergoes an adiabtic process, thus
pV = nRT and dq = 0.
Further, it is given that the process is reversible so that pex = pgas = p.
To show the validity of the given equation above requires the derivation of the
equations needed to describe an adiabatic process of an ideal gas. This can be
done in various manners.
Manner 1:
One starts with the first law of thermodynamics: dU = dq + dw.
In a next step it is evaluated how the different terms can be computed. First, the
derivation of the equation to describe the work. In general only pV work is taken
into account so that dw = - pexdV.
The process is reversible so that the external pressure is always equal to the
pressure of the system: dw = - p dV
For an adiabatic pressure no heat is exchanged between the system and the
surroundings: dq = 0 leading to dU = dw = -p dV
With this the change of the internal energy can be described. However, to get the
equation given above we need to combine this description with the description of
the change of the internal energy based on U = U(V,T)
 ∂U 
 ∂U 
dU = 
 dT + 
 dV . Further it is known that the internal energy of an
 ∂T V
 ∂V T
ideal gas only changes with a change of the temperature. In other words, we know
that at constant temperature:
 ∂U 

 =0
 ∂V T
Generally, the partial derivative of the internal energy with respect to the
temperature at constant volume is equal to the heat capacity at constant volume:
 ∂U 
 ∂U 

 = CV = n ⋅ CV , giving: dU = 
 dT = CV dT = n ⋅ CV dT
 ∂T V
 ∂T V
The two manners to describe the change of the internal energy can be equalized
giving:
− pdV = n ⋅ CV dT
In order to get the desired equation we need to further modify the equation. In the
following the initial state is denoted by ‘1’, the final state by ‘2’.
First the relation between T and V is introduced before the relation between p and
T can be given. For this purpose first p is replaced by the ideal gas law, namely by
nRT/V:
nRT
−
dV = n ⋅ CV dT
V
Then the variables are split, meaning all volumes are brought to one side and all T
to the other side, leading to:
nRT
dV
dT
−
dV = n ⋅ CV dT ⇒ − R
= CV
V
V
T
Assuming that CV is constant, the last equation can be integrated between the
initial and final state:
V2
∫ ( −R )
V1
V
T
T
2
2
dV
dV 2
dT
dT
= −R ∫
= ∫ CV
= CV ∫
V
V
T
T
V1
T1
T1
V 
V 
T 
− R ln  2  = R ln  1  = CV ln  2 
 V1 
 V2 
 T1 
R
CV
R
CV
CV
V 
T 
V 
T 
V  T  R
ln  1  = ln  2  ⇒  1  =  2  ⇒  1  =  2 
 V2 
 T1 
 V2 
 T1 
 V2   T1 
With the last equation at the right, the relation between T and V for an ideal gas
undergoing an adiabatic process is given. Based on this equation the relationship
between p and T can be derived by replacing the volumes, V1 en V2, by the ideal
gas law, leading to:
 nRT1 
CV
 V1   p1   T1 p2   T1   p2   T2  R
  =  nRT  = 
 =  ⋅  =  
2 
 V2  
 T2 p1   T2   p1   T1 
 p 
 2 
CV
 p  T  R
⇒ 2 = 2 
 p1   T1 
−
CV
T  T  R
⋅ 2  =  2 
 T1   T1 
+1
T 
= 2 
 T1 
CV + R
R
T 
= 1 
 T2 
−
CV + R
R
R
 T   p  CV + R
⇒ 1 = 2 
 T2   p1 
To obtain the above given equation the exponent needs to be rewritten. Therefore,
we recall that for an ideal gas CP = CV + nR and also
CP = nCP = CV + nR = nCV + nR ⇒ CP = CV + R
Incorporating this into the equation leads to the wanted expression
 T1   p2 
 = 
 T2   p1 
p 
= 2 
 p1 
−
R
CV + R
CV CP
−
CV CV
CP
CV
p 
= 2 
 p1 
−
p 
= 2 
 p1 
T   p 
⇒ 1 = 2 
 T2   p1 
CP −CV
CP
p 
= 2 
 p1 
CV − CP
CP
p 
= 2 
 p1 
CV − CP CV
⋅
CP CV
1−γ
γ
1−γ
γ
Manner 2:
Also for this approach we start with the first law of thermodynamics
dU = dq + dw.
However, because the relationship between T and p is wanted, we approach the
problem via the enthalpy because the enthalpy is commonly defined in the
temperature, pressure space, H = H(T,p). The first law of thermodynamics and the
enthalpy are related via the definition of the enthalpy H = U + pV and its total
differential dH = dU + pdV + Vdp. Incorporating the first law while taking into
account that for the given process dq = 0 and dw = - pdV (see also manner 1)
gives:
dH = -pdV + pdV + V dp = Vdp
With this equation the change of the enthalpy for an reversible adiabatic process
of an ideal gas can be described. Next the change of the enthalpy is also described
based on the fact that H = H(p,T). This gives another expression of the total
differential of the enthalpy
 ∂H 
 ∂H 
dH = 
 dp .
 dT + 
 ∂T  p
 ∂p T
Accounting for the fact that the enthalpy of a system consisting of an ideal gas
does not change if the temperature is kept constant gives:
 ∂H 

 =0
 ∂p T
Further we know that in general
 ∂H 

 = CP = n ⋅ CP . This gives for the total differential of the enthalpy of an
 ∂T  P
 ∂H 
ideal gas: dH = 
 dT = CP dT = n ⋅ CP dT
 ∂T  P
This gives with the other equation describing dH:
Vdp = n ⋅ CP dT
For the further derivation we describe the initial state by the index ‘1’ and the end
state by ‘2’. Next the number of parameters varying during the process is reduced
by describing the volume by the ideal gas law V = nRT/p:
nRT
dp = n ⋅ CP dT
p
Before this equation can be integrated from its initial state ot its final state, the
parameters are split:
nRT
dp
dT
dp = n ⋅ CP dT ⇒ R
= CP
p
p
T
When keeping CP constant the equation can be integrated giving:
p2
∫R
p1
p
T
T
2
2
dp
dp 2
dT
dT
= R∫
= ∫ CP
= CP ∫
p
p T1
T
T
p1
T1
p 
T 
R ln  2  = CP ln  2 
 p1 
 T1 
R
p 
T 
ln  2  = ln  2 
 p1 
 T1 
CP
R
 p  T 
⇒ 2  = 2 
 p1   T1 
CP
R
 T   p  CP
⇒ 2 = 2 
 T1   p1 
Next, we introduce for the ideal gas CP = CV + nR or more precisely
CP = nCP = CV + nR = nCV + nR ⇒ CP = CV + R
Leading finally to the wanted expression:
−1
 T2   T1 
 p2 
 =  = 
 T1   T2 
 p1 
p 
= 2 
 p1 
CV CP
−
CV CV
CP
CV
p 
= 2 
 p1 
T   p 
⇒ 1 = 2 
 T2   p1 
3.
T   p 
⇒ 1 = 2 
 T2   p1 
−
CP − CV
CP
p 
= 2 
 p1 
CV −C P
CP
p 
= 2 
 p1 
CV − CP CV
⋅
CP CV
1−γ
γ
1−γ
γ
In the last lectures you got acquainted with a number of thermodynamic properties. For some
specific processes some of the equations can be simplified. Below a number of equations are
given. Please specify for each of these equations if they are generally valid or if they are only valid
for specific cases, e.g., only valid for an ideal gas or for an isobaric process. Please state for each
of these equations for which cases they are valid. Further, state if the equation is integrated.
a. ∆U = q + w → first law of thermoydnamics; integrated; general valid
b. dU = dq + dw → first law of thermodynamics; differential form; general valid
c. q = ∆H → integrated; only valid for either an isobaric and reversible process or an
isochoric and reversible process
∆H
→ integrated; CP=const; isobaric process of any system or a not isothermal
∆T
d.
CP =
e.
process of a system consisting of ani deal gas
dH = dU + d (n ⋅ R ⋅ T ) → differential form; general valid for a system consisting of
an ideal gas
g.

n2 ⋅ a 
p
+

 ⋅ (V − nb ) = n ⋅ R ⋅ T → equation of state of a real gas
V2 

w = − pex ⋅ ∆V → integrated; general description of work done during an irreversible
h.
i.
process
(∆U)T = 0 → integrated; isothermal process of an ideal gas
dw = − pgas ⋅ dV → differential form; reversible process of a system consisting of a gas
f.
4.
CP − CV
CP
1.0 kg water is evaporated at 100o C and a constant pressure of 0.101 MPa. Compute ∆U for the
process if the required heat to evaporate 1.0 kg (liquid) water is 2260 kJ. The volume of 1kg (liquid)
water is 10-3 m3, of 1 kg steam 1.67 m3.
Solution:
The system consists of 1 kg water (mH2O = 1.0 kg). The initial liquid water is isobarically and
isothermally transformed from liquid water to stteam (dT= 0, dp =0). The initial state of the
process of th einitial liquid water is described by 100oC and 0.101 MPa; the final state of the steam
by 100oC and 0.101 MPa.
Further, it is given that for the evaopration of 1 kg (liquid) water
qvap= 2260kJ needs to be added to the system and the absolute volume of the liquid water at these
conditions is VL=10-3 m3 and of the steam is VV=1.67 m3.
Wanted: ∆Uvap:
Manner 1:
Start with first law of thermodyanmics: dU = dq + dw
The work can be conputed by: dw = - pex·dV and thus for the given process:
w = - pex·∆V = - pex·(VV – VL) = - 0.101 MPa·(1.67 – 10-3)m3 = - 0.101 106 Pa·(1.67 – 10-3)m3=0.101 106 N m-2·(1.67 – 10-3)m3 ⇒ w = -168569 J
The exchanged heat to evaporate the water is given: q = 2260000 J. Thus the change of the
internal energy can be computed:
dU = dq + dw ⇒ ∫ dU = ∫ dq + ∫ dw
⇒ ∆U = q + w = 2260000 J − 168569 J = 2091431J = 2091.43kJ
Manner 2:
For an isobaric process the heat exchanged is equal to the enthapy change of the system:
H = U + pV, so that U = H – pV and dU= dH – d(pV) = dH – pdV – Vdp. For an isobaric process
(dp =0): dU = dH – pdV.
Integration of this equation from the initial to the end state gives:
∆Uvap= ∆Hvap - p·∆Vvap.
∆Hvap and ∆Vvap= VV-VL are given so that:
∆Uvap=∆Hvap-p·∆Vvap=∆Hvap–p·(VV-VL)
=2260kJ–0.101*106Pa *(1.67 – 10-3) m3
= 2260*1000 J–0.101*106 N m-2 *(1.67 – 10-3) m3
= 2091431 J = 2091.431 kJ
Thus we can see that the change of the internal energy when completely evaporating 1kg (liquid)
water at 1000C and 0.101 MPa at constant temperature and pressure is 2091.43 kJ.
5.
Show that for a closed p,V,T- system the following equations are valid:
CV ⋅ dT
for an isochoric process
T
C ⋅ dT
b) d S = P
for an isobaric process.
T
a)
dS =
Solution:
The expressions given in the exercise are both based on the following expression:
dS =
dqrev
. Thus by means of the first law of thermodynamics and the definition of the enthalpy
T
these expressions can be derived. For the derivation of both expression, an reversible process is
assumed: pex = psysteem = p
Ad a) Besides the fact that the process is reversible, the index already indicates here an isochoric
process is assumed (dV = 0). We start with the first law of thermodynamics: dU = dqrev + dwrev.
The work done can be computed by: dw = - pex dV = - p dV. And for an isochoric process: dwrev =
0 and wrev = 0. And thus: dU = dqrev. The change of the internal energy dU can be described by:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 dV + 
 dT
 ∂V T
 ∂T V
 ∂U 
dV = 0 ⇒ dU = 
 dT = CV dT
 ∂T V
dqrev CV ⋅ dT
So that dqrev=dU = CV dT and further: dS =
=
T
T
qed
Ad b) Besides the fact that the process is reversible, indicates the index p that an isobaric process
(dp=0) needs to be described. We start again with the first law of thermodynamics: dU = dqrev +
dwrev. The work is computed by dw = - pex dV = - p dV. Because the process is isobaric, it is more
practical to work with the enthalpy than with the internal energy. The change of the enthalpy dH
can be described by:
 ∂H 
 ∂H 
H = H ( p, T ) ⇒ dH = 
 dp + 
 dT
 ∂T  P
 ∂p T
 ∂H 
dp = 0 ⇒ dH = 
 dT = CP dT
 ∂T  p
The relation between the internal energy, enthalpy, the work and the heat can be derived from the
definition of the enthalpy:
H ≡ U + p ⋅V ⇒ dH = dU + d ( p ⋅V ) = dU + p ⋅ dV + V ⋅ dp 
 ⇒ dH = dqrev + V ⋅ dp
dU = dqrev + dwrev = dqrev − p ⋅ dV

dp = 0 ⇒ dH = dqrev
dqrev CP ⋅ dT
qed
Thus dqrev=dH = CP dT and: dS =
=
T
T
6.
1 mol of an ideal gas is heated and compressed from 250C and 1 bar to 3000C and 10 bar. For this
process a cylinder with a frictionless piston is used. The process is done in three different ways. Show
that ∆U, ∆H and ∆S are state functions while q and w not. For this purpose compute for the three
different ways (see below) ∆U, ∆H, ∆S, q and w.
a.
First step is isothermal compression to 10 bar followed by isobaric heating to 3000C.
b.
First step is isobaric heating to 3000C followed by isothermal compression to 10 bar.
c.
First step is compression described by
p ⋅ V γ = constant with γ =
CP
followed by
CV
isobaric heating or cooling (depending on the reached temperature after the first
process step) to 3000C.
Further, it is given that
CP =38 J mol-1 K-1 = constant
Solution:
The system consist of 1mol ideal gas so that the ideal gas law is valid: pV = nRT.
The prupose of this exercise is to see that q and w depend on the chosen path of the process while U, H and
S are independent of the chosen path. Thus for each of the described processes q, w, ∆U, ∆H and ∆S are
computed and then compared. The initial and final states for all the processes are the same so that the
changes of the path-independent properties should be the same. All the processes describe the following
change:
1 mol ideal gas ( 250C, 1 bar) → 1 mol ideal gas ( 3000C, 10 bar)
The process is divided into two steps. Thus the changes of the properties need to be summed up to get the
complete change. All processes are done in a cylinder with a frictionless piston, meaning that all processes
are reversible, pex = pgas = p.
a)
Step I: isothermal compression:
1 mol ideal gas (250C, 1bar)→
→ 1mol ideal gas (250C, 10bar)
Step II: isobaric heating
1 mol ideal gas (250C, 10bar) → 1 mol ideal gas (3000C, 10bar)
Step I: dT = 0; T1 = T2 = 298,15 K; p1 = 1 bar; p2 = 10 bar
Reversible→pex = pgas = p
We start again with the first law of thermodynamics:
dU = dq + dw
Followed by the description of dw:
dw = - pex·dV = -p·dV.
During the isothermal compression, the pressure and the volume of the system changes so that before
integration the equation needs to be rewritten in terms of temperature and either the volume or the pressure.
For this purpose, the ideal gas law is incorporated and the pressure is replaced by the ideal gas law: p =
nRT/V
dw = − pdV = −
nRT
dV
V
V2
V
2
dV
 nRT 
⇒ w = ∫ dw = ∫  −
 dV = −nRT ∫
V 
V
V1 
V1
V 
= −nRT ln  2 
 V1 
V2 and V1 are not given explicitly but can be obtained with the given data and the ideal gas law (V =
nRT/p):


waI = − nRT ln 





p 
 = −nRT ln  1 

 p2 


J
 1bar 
= −1mol ⋅ 8.314
⋅ 298.15 K ln 

molK
 10bar 
= 5707.69 J
nRT
p2
nRT
p1
Next, ∆U is computed (Tip: fora n isothermal process first dU or dH should be computed before q is
computed):
U = U(V,T) and
 ∂U 
 ∂U 
dU = 
 dT + 
 dV .
 ∂T V
 ∂V T
For ani deal gas, the internal energy only changes if the temperature changes during the process:
 ∂U 

 =0
 ∂V T
Further, in general:
 ∂U 

 = CV = n ⋅ CV , Leading to:
 ∂T V
 ∂U 
dU = 
 dT = CV dT = n ⋅ CV dT
 ∂T V
T2
(T1 = T2)
⇒ ∆U aI = ∫ n ⋅ CV dT = 0
T1
And with the first law of thermodynamics, we get for q:
dU = dq + dw ⇒ dq = dU – dw ⇒ q = ∆U – w = 0 – 5707.69 J
qaI = -5707.69 J
Next, ∆H needs tob e computed. This can be done in two ways:
1) via the definition of the enthalpy: H = U + pV
2) via H = H(p,T) and
 ∂H 
 ∂H 
dH = 
 dp
 dT + 
 ∂T  p
 ∂p T
Ad 1) H = U + pV and thus dH = dU + d(pV) = dU + d(nRT) ⇒ ∆HaI= ∆U + ∆(nRT) = 0 + 0 = 0J.
Ad 2) The enthalpy of an ideal gas only changes with changes of the temperature:
 ∂H 

 =0
 ∂p T
Further in general we know that
 ∂H 

 = CP = n ⋅ CP , giving:
 ∂T  P
 ∂H 
dH = 
 dT = CP dT = n ⋅ CP dT
 ∂T  P
T2
T2
T1
T1
(T1 = T2)
⇒ ∆H aI = ∫ n ⋅ CP dT = n ⋅ CP ∫ dT = 0
Last the change of the entropy needs to be computed. In general the change of the entropy is computed by:
dS =
dqrev
dq
q
1
. For an isothermal process: ∆S = ∫ dS = ∫ rev = ∫ dqrev = rev
T
T
T
T
The real process is also reversible so that the computed value of the exchanged heat can be filled into the
equation:
∆S aI =
qrev −5707.69 J
J
=
= −19.15
T
298 K
K
Step II: dp = 0; p1 = p2 = 10 bar; T1 = 298.15 K; T2 = 573.15K
reversible→pex = pgas = p
Again we start with the first law of thermodynamics: dU = dq + dw followed bythe computation of dw:
dw = -pex dV = - pdV, here the pressure is constant so that:
dw = − pdV ⇒ w =
V2
V2
V1
V1
∫ ( − p ) dV = − p ∫ dV = − p (V
2
− V1 )
The values for the volumes are not given explicitly but can be obtained replacing the volume by the ideal
gas law V = nRT/p:
 nRT2 nRT1 
w = − p (V2 − V1 ) = − p 
−
 = −nR (T2 − T1 )
p 
 p
J
w = −1mol ⋅ 8.314
( 573.15 − 298.15) K
molK
waII = −2286.35 J
Because the system consists of ani deal gas, it does not matter whether first the change of the internal
energy or of the enthalpy is calculated. Here the different approaches are given:
Manner I: We start with U = U(V,T) and
 ∂U 
 ∂U 
dU = 
 dT + 
 dV . For ani deal gas the total differential reduces to (see for the derivation
 ∂T V
 ∂V T
step I):
 ∂U 
 ∂U 
dU = 
 dT + 
 dV = CV dT = n ⋅ CV dT
 ∂ T V
 ∂ V T
⇒ ∆U aII =
T2 = 573.15 K
∫
n ⋅ CV dT = n ⋅ CV
T1 = 298.15 K
T2 = 573.15 K
∫
dT = n ⋅ CV ⋅ (T2 − T1 )
T1 = 298.15 K
Before the actual value for ∆Ucan be computed, the value for CV needs to be known. In the exercise only
the value for CP is given. However, for an ideal gas: CP = CV + n R so that: CV = CP – nR or
CV = n ⋅ CV = CP − n ⋅ R = n ⋅ C P − n ⋅ R ⇒ CV = C P − R .
Incorporating this in the expression to compute the change of the internal energy gives:
(
)
∆U aII = n ⋅ CV ⋅ (T2 − T1 ) = n ⋅ CP − R ⋅ (T2 − T1 )
= 1mol ⋅ ( 38 − 8.314 )
J
⋅ ( 573.15 − 298.15 ) K
molK
= 8163.65 J
The heat exchanged is computed by:
dU = dq + dw ⇒ dq = dU – dw ⇒ q = ∆U – w = 8163.65 J – (-2286.35) J
qaII = 10450 J
For the computation of ∆H we start with H = U + pV and
dH = dU + d(pV) = dU + d(nRT) = dU + nR dT
which gives after integration:
∆HaII = ∆UaII + nR ∆TaII = 8163.65 J + 1mol ⋅ 8.314 J mol-1 K-1 ⋅ (573.15 – 298.15) K
∆HaII = 10450 J
Manner II: For this approach we start with H and the relation between the enthalpy and the first law of
thermodynamics:
H = U + pV, and dH= dU + d(pV) = dU + pdV + Vdp. The process is isobaric dp = 0 and dU = dq –
pdV, and that the process is reversible. Incoporating this into the equations gives:
dH = dq – pdV + pdV = dq. Further, we know that H = H(p,T) and
 ∂H 
 ∂H 
dH = 
 dp
 dT + 
 ∂T  p
 ∂p T
The enthalpy and the internal energy of an ideal gas only change if the temperature changes during a
process:
 ∂H 

 =0
 ∂p T
In general we know that
 ∂H 

 = CP = n ⋅ CP , so that:
 ∂T  P
∂H 
dH = 
 dT = CP dT = n ⋅ CP dT
 ∂ T P
⇒ ∆H aII =
T2 =573.15 K
∫
T1 = 298.15 K
∆H aII = 1mol ⋅ 38
n ⋅ CP dT = n ⋅ CP
T2 = 573.15 K
∫
dT = n ⋅ CP ⋅ (T2 − T1 )
T1 = 298.15 K
J
⋅ ( 573.15 − 298.15 ) K = 10450 J
molK
From this it can be deduced that qaII= 10450 J because qaII= ∆HaII. The change of the internal energy
follows from the first law of thermodynamics:
dU= dq + dw ⇒ ∆UaII = qaII + waII = 10450 J – 2286.35 = 8136.65 J
Last the change of the entropy is computed starting with:
dS =
dqrev
. Thus a reversible and isobaric
T
process needs to be described. Above we have shown that dqrev = dH giving:
dqrev
dH
c ⋅ dT
=∫
=∫ P
. If it can be assumed that Cp = constantwe get:
T
T
T
T 
dT
∆S = cP ⋅ ∫
= cP ⋅ ln  2  and for the given process:
T
 T1 
∆S = ∫ dS = ∫
T 
T 
J
 573K 
∆S aII = cP ⋅ ln  2  = n ⋅ cP ⋅ ln  2  = 1mol ⋅ 38
⋅ ln 

mol ⋅ K
 298 K 
 T1 
 T1 
J
∆S aII = 24.84
K
Now all the properties for the complete process are computed:
∆Ua= ∆UaI + ∆UaII = 0 + 8136.65 J = 8136.65 J
∆Ha= ∆HaI + ∆HaII = 0 + 10450 J = 10450 J
∆Sa= ∆SaI + ∆SaII = -19.15 J K-1 + 24.84 J K-1 = 5.69 J K-1
qa = qaI + qaII = = -5707.7 J + 10450 J = 4742.3 J
wa = waI + waII = 5707.7 J – 2286.65 J = 3421.35 J
b)
Step I: isobaric heating:
1 mol ideal gas (250C, 1 bar) → 1 mol ideal gas (3000C, 1 bar)
Step II: isothermal compression:
1mol ideal gas (3000C,1bar) → 1mol ideal gas (3000C, 10bar)
Al required equation are already derived in exercise a. Just as mentioned in exercise a some of the changes
of the properties can be comouted in different ways. In this part the equations are used and not dervied
again. So for the derivation, please check exercise a. Thereby, note that the equations needed for step 1 oif
exercise b are derived for step II of exercise a. The equations for the description of step II of exercise b are
the same equations as for step I of exercise a.
Step I: dp = 0; p1 = p2 = 1 bar; T1 = 298.15 K; T2 = 573.15K
reversible→pex = pgas = p
dw = − pdV ⇒ w =
V2
V2
V1
V1
∫ ( − p ) dV = − p ∫ dV = − p (V2 − V1 )
Note: process is isobaric.
 nRT2 nRT1 
J
w = −p
−
⋅ ( 573.15 − 298.15 ) K
 = −nR (T2 − T1 ) = −1mol ⋅ 8.314
p 
molK
 p
wbI = −2286.35 J
For an isobaric process q=∆H
∂ H 
dH = 
 dT = CP dT = n ⋅ CP dT
 ∂ T P
⇒ ∆H bI =
T2 =573.15 K
∫
n ⋅ CP dT = n ⋅ CP
T1 = 298.15 K
∆H bI = 1mol ⋅ 38
T2 =573.15 K
∫
dT = n ⋅ CP ⋅ (T2 − T1 )
T1 = 298.15 K
J
⋅ ( 573.15 − 298.15 ) K = 10450 J = qbI
molK
dU= dq + dw ⇒ ∆UbI = qbI + wbI = 10450 J – 2286.35 = 8136.65 J
dqrev
dq
c ⋅ dT
dH
and because process is isobaric dqrev = dH: ∆S = ∫ dS = ∫ rev = ∫
. For
=∫ P
T
T
T
T
T 
dT
Cp = constant: ∆S = cP ⋅ ∫
= cP ⋅ ln  2  and thus:
T
 T1 
dS =
T 
T 
J
 573K 
∆SbI = cP ⋅ ln  2  = n ⋅ cP ⋅ ln  2  = 1mol ⋅ 38
⋅ ln 

mol ⋅ K
 298 K 
 T1 
 T1 
J
∆SbI = 24.84
K
Step II: dT = 0; T1 = T2 = 573.15 K; pex = pgas = p; p1 = 1 bar; p2 = 10 bar
reversible→pex = pgas = p
dU = dq + dw
and
dw = - pexdV = -pdV.
and
V 
w = − nRT ln 2 
 V1 
And with the ideal gas law:
p 
w = −nRT ln  1 
 p2 
wbII = −1mol ⋅ 8.314
J
 1bar 
⋅ 573.15 K ln 

molK
 10bar 
= 10972.2 J
 ∂U 
 ∂U 
dU = 
 dT + 
 dV = CV dT = n ⋅ CV dT
 ∂ T V
 ∂ V T
⇒ ∆U bII =
T2 =573.15 K
∫
(T1 = T2)
n ⋅ CV dT = 0 J
T1 =573.15 K
And q:
dU = dq + dw ⇒ dq = dU – dw ⇒ q = ∆E – w = 0 – 10972.2 J
qbII = – 10972.2 J
H = U + pV and dH = dU + d(pV) = dU + d(nRT) ⇒ ∆HbII= ∆U + ∆(nRT) = 0 + 0 = 0 J
dS =
dqrev
dq
q
1
Process is isothermal so that: ∆S = ∫ dS = ∫ rev = ∫ dqrev = rev
T
T
T
T
∆S aI =
qrev −10972.2 J
J
=
= −19.15
T
573K
K
With this we get:
∆Ub= ∆UbI + ∆UbII = 8136.65 J+ 0 = 8136.65 J
∆Hb= ∆HbI + ∆HbII = 10450 J + 0 = 10450 J
∆Sb= ∆SbI + ∆SbII = 24.84 J K-1 -19.15 J K-1 = 5.69 J K-1
qb = qbI + qbII = = 10450 J – 10972.2 J = -522.21 J
wb = wbI + wbII = -2286.65 J + 10972.2 = 8685.86 J
c)
Step I: compression with pVγ= const and
γ =
CP
CV
1 mol ideal gas (250C, 1 bar) → 1 mol ideal gas (T2, 10 bar)
Step II: isobaric heating/cooling:
1mol ideal gas (T2, 10 bar) → 1mol ideal gas (3000C, 10bar)
Step I: pVγ= const with
γ =
CP CP
; T1 = 573.15 K; p1 = 1 bar; p2 = 10 bar
=
CV CV
reversible→pex = pgas = p
We start with the first law of thermodynamics: dU = dq + dw.
For a reversible process dw = - p dV. This equation cannot be integrated because p and V change during the
process. Therefore, the pressure is replaced by the ideal gas law (p = nRT/V). However, this does not help
as also the temperature changes during the process. For the process it is given that pVγ= const = C (we just
call the constant C), and replace the pressure p by CV-γ:
dw = − pdV = −
⇒ w = ∫ dw =
C
dV = −C ⋅ V − γ dV
γ
V
V2
V1
w=−
−C
V21− γ − V11− γ
+1
γ
∫ (−C ⋅V )dV = −γ
−
(
)
C  V2 V1 
− 

1 − γ  V2γ V1γ 
There are no explicit values given for the volumes at the initia land the end state. Also the value for the
constant C is not given so that the volume Vγ is replaced by C/p:




C  V2 V1 
C  V2 V1 
1
w=−
−
=−
 γ − γ=−
(V2 ⋅ p2 − V1 ⋅ p1 )
C
1 − γ  V2 V1 
1− γ  C
1− γ


 p2 p1 
To allow the computation of the work, the product Vp needs to be replaced by the ideal gas law: pV = n RT
w=−
1
nR
nRT2 − nRT1 ) = −
(
(T2 − T1 )
1− γ
1− γ
For the computation of T2 we use pVγ= const. => p1V1γ= p2V2γ. And incorporating the ideal gas law:
γ
γ
 nRT1 
 nRT2 
p
p
γ
γ
γ
γ
p V = p V ⇒ p1 
 = p2 
 ⇒ γ1 ⋅ ( nR) ⋅ T1 = γ2 ⋅ ( nR) ⋅ T2
p1
p2
 p1 
 p2 
γ
1 1
γ
2 2
1− γ
1
⇒p
γ
1
1− γ
2
⋅T = p
p 
⇒ T2 =  1 
 p2 
1− γ
γ
p 
p1− γ
⋅ T ⇒ T = 11− γ ⋅ T1γ =  1 
p2
 p2 
γ
2
1− γ
γ
2
⋅ T1γ
⋅ T1
For the final computation of the temperature , we first compute γ:
γ=
J
molK
CP n ⋅ C P C P
CP
=
=
=
=
CV n ⋅ CV CV CP − R
( 38 − 8.314 )
1− γ
γ
38
J
molK
= 1.28006
= −0.2818
1−γ
p γ
 1bar 
T2 =  1  ⋅ T1 = 

 10bar 
 p2 
T2 = 493.43K
−0.2188
⋅ 298.15 K
And;
nR
w=−
(T2 − T1 ) = −
1− γ
w = 5797.08 J
J
molK ⋅ ( 493.43 − 298.15 ) K
1 − 1.28006
1mol ⋅ 8.314
 ∂U 
 ∂U 
dU = 
 dT + 
 dV = CV dT = n ⋅ CV dT
 ∂ T V
 ∂ V T
⇒ ∆U cI =
T2 = 493.43 K
∫
n ⋅ CV dT = n ⋅ CV
T1 = 298.15 K
(
T2 = 493.43 K
∫
dT = n ⋅ CV ⋅ (T2 − T1 )
T1 = 298.15 K
)
= n ⋅ CP − R ⋅ (T2 − T1 ) = 1mol ⋅ ( 38 − 8.314 )
J
⋅ ( 493.43 − 298.15 ) K
molK
= 5797.08 J
dU = dq + dw ⇒ dq = dU – dw ⇒ q = ∆U – w = 5797.08 J – 5797.08 J
qcI = 0 J
From this we can see that pVγ = const actually means that the process is adiabatic (dq=0).
For an ideal gas:
∂ H 
dH = 
 dT = CP dT = n ⋅ CP dT
 ∂ T P
⇒ ∆H cI =
T2 = 493.43 K
∫
n ⋅ CP dT = n ⋅ CP
T1 = 298.15 K
∆H cI = 1mol ⋅ 38
T2 = 493.43 K
∫
dT = n ⋅ CP ⋅ (T2 − T1 )
T1 = 298.15 K
J
⋅ ( 493.43 − 298.15 ) K = 7420.64 J
molK
dqrev
with the knowledge that the process is reversible and adiabtic (qrev = 0J):
T
dq
J
∆ScI = ∫ dS = ∫ rev = 0
T
K
And for
dS =
Step II: T2 = 493.43 K. The second process step is thus an isobaric and reversible heating: dp = 0; p1
= p2 = 1 bar; T1 = 493.43 K; T2 = 573.15K
reversible→pex = pgas = p
Here we can use the same equations as the equations derived for exercise a step II or exercise b step I:
dw = − pdV ⇒ w =
V2
V2
V1
V1
∫ ( − p ) dV = − p ∫ dV = − p (V
2
− V1 )
With V = nRT/p:
w = − nR (T2 − T1 ) = −1mol ⋅ 8.314
J
⋅ ( 573.15 − 493.43) K
molK
wcII = −662.79 J
∂ H 
dH = 
 dT = CP dT = n ⋅ CP dT
 ∂ T P
⇒ ∆H cII =
T2 = 573.15 K
∫
T1 = 493.43 K
∆H cII = 1mol ⋅ 38
n ⋅ CP dT = n ⋅ CP
T2 =573.15 K
∫
dT = n ⋅ CP ⋅ (T2 − T1 )
T1 = 493.43 K
J
⋅ ( 573.15 − 493.43) K = 3029.36 J
molK
Because it is an isobaric process qcII= ∆HcII = 3029.36 J.
dU= dq + dw ⇒ ∆UcII = qcII + wcII = 3029.36 J – 662.79 J = 2366.57 J.
dqrev
dq
c ⋅ dT
dH
with dqrev = dH (because isobaric): ∆S = ∫ dS = ∫ rev = ∫
.
=∫ P
T
T
T
T
T 
dT
With Cp = constant: ∆S = cP ⋅ ∫
= cP ⋅ ln  2 
T
 T1 
dS =
T 
T 
J
 573K 
∆ScII = cP ⋅ ln  2  = n ⋅ cP ⋅ ln  2  = 1mol ⋅ 38
⋅ ln 

mol ⋅ K
 493.43K 
 T1 
 T1 
J
∆ScII = 5.69
K
And for the complete process:
∆Uc= ∆UcI + ∆UcII = 5797.08 J + 2366.57 J = 8136.65 J
∆Hc= ∆HcI + ∆HcII = 7420.64 J + 3029.36 J = 10450 J
∆Sc= ∆ScI + ∆ScII = 0 J K-1 + 5.69 J K-1 = 5.69 J K-1
qc = qcI + qcII = = 0 J + 3029.36 J = 3029.36 J
wc = wcI + wcII = 5797.08 J – 662.79 J = 5134.29 J
a
b
c
∆E [J]
8136.65
8136.65
8136.65
∆H [J]
10450
10450
10450
∆S [J K-1]
5.69
5.69
5.69
q [J]
4742.3
-522.21
3029.36
w [J]
3421.35
8685.86
5134.29
From this table it can be seen that U, H and S independent of the path is (same values for all paths) while q
and w depend on the path (different values for the path).
7.
a) Show that:
 ∂G 
 ∂G 
 =V

 = −S & 
∂
T

p
 ∂p T
The Gibbs energy is defined as G ≡ H – T S = U + pV – TS Further the Gibbs energy can be
expressed in terms of temperature and pressure: G = G(p, T). The total differential of the Gibbs
energy can be expressed based on these two equations, leading to the given expressions:
G ≡ H − TS = U + pV − TS ⇒ dG = dU + d ( pV ) − d (TS ) = dU + pdV + Vdp − TdS − SdT
dU = TdS − pdV ⇒ dG = TdS − pdV + pdV + Vdp − TdS − SdT ⇒
dG = Vdp − SdT


 ∂G 
 ∂G 

G = G (T , p ) ⇒ dG = 
 dp 
 dT + 
 ∂T  p
 ∂p T 

 ∂G 
 ∂G 
( dG )T = 
 dp = Vdp ⇒ 
 =V

 ∂p T
 ∂p T
⇒
( dG ) =  ∂G  dT = − SdT ⇒  ∂G  = − S




p

 ∂T  p
 ∂T  p

b) Show that:
 ∂U 

 = −p &
 ∂ V S
 ∂U 

 =T
 ∂ S V
We know that dU = T dS – p dV and that U = U(S, V). In this way, the total differential of the
internal energy can be described in two different ways:
dU = TdS − pdV


 ∂U 
 ∂U 

U = U ( S , V ) ⇒ dU = 
dS
+
dV




 ∂S V
 ∂V  S


 ∂U 
 ∂U 
( dU )V =  ∂S  dS = TdS ⇒  ∂S  = T

V

V

⇒
( dU ) =  ∂U  dV = − pdV ⇒  ∂U  = − p




S
 ∂V  S
 ∂V  S

c) Show that:
∂ H 

 =T
 ∂ S p
&
∂ H 

 =V
 ∂ p S
The enthalpy is defined as H ≡ U + pV. However, the enthalpy H can also be expressed in terms
of S and p, H = H(S, p). Based on these two equations the total differential can be written.
Combination gives the desired equations:
H ≡ U + pV ⇒ dH = dU + d ( pV ) = dU + pdV + Vdp
dU = TdS − pdV ⇒ dH = TdS − pdV + pdV + Vdp ⇒
dH = TdS + Vdp


 ∂H 
 ∂H 

H = H ( S , p ) ⇒ dH = 
dS
dp
+




 ∂S  p
 ∂p  S 

 ∂H 
 ∂H 
( dH ) p =  ∂S  dS = TdS ⇒  ∂S  = T

p

p

⇒
( dH ) =  ∂H  dp = Vdp ⇒  ∂H  = V




S

 ∂p  S
 ∂p  S

∂ A

 = −p &
 ∂ V T
d)Show that:
∂ A

 = −S
 ∂ T V
We know that A ≡ U – T S and that A = A(V, T). With this the total differential of the Helmholtz
energy can be described in two ways:
A ≡ U − TS ⇒ dA = dU − d (TS ) = dU − TdS − SdT
dU = TdS − pdV ⇒ dA = TdS − pdV − TdS − SdT ⇒
dA = − pdV − SdT


 ∂A 
 ∂A 

A = A (T , V ) ⇒ dA = 
 dT + 
 dV 
 ∂T V
 ∂V T


 ∂A 
 ∂A 
( dA)V =  ∂T  dT = − SdT ⇒  ∂T  = − S

V

V

⇒
( dA) =  ∂A  dV = − pdV ⇒  ∂A  = − p




T

 ∂V T
 ∂V T
8) Show that:
a)
 ∂T 
∂ p

 = −

 ∂ V S
 ∂ S V
For the derivation of the expression, first it needs to be clarified which first derivative of a state
function, defined in the S, V-space, is equal to T and p. From the first exercise we know that U =
U(S, V) and
and
With this we get
dU = TdS − pdV
 ∂U 
 ∂U 
U = U (V , S ) ⇒ dU = 
 dV + 
 dS
 ∂ V S
 ∂ S V
 ∂U 
−p = 

 ∂ V S
&
 ∂U 
T =

 ∂ S V
U is homogeneous and continuous function so that the rule of Schwartzcan be applied leading to:

∂
∂ p

−
 =
 ∂ S V 


  ∂U   
 ∂U   
 ∂ 

 
 
2
2
  ∂ V  S   = ∂ U = ∂ U =    ∂ S V   =  ∂ T 

  ∂ V  S
∂S
∂ V ∂ S ∂ S∂ V 
∂V



V

S
b)
 ∂V   ∂ T 


 =
 ∂ S  p  ∂ p S
For this derivation first it needs to be shown of which state function in the S,p-space the first
derivatives are equal to V and T. from exercise 1 we know that H = H(p, S) and that:
dH = TdS + Vdp
Based on the total differential we get:
∂ H 
∂ H 
H = H ( S , p ) ⇒ dH = 
 dp
 dS + 
 ∂ S p
 ∂ p S
And thus:
∂ H 
T =

 ∂ S p
&
∂H 
V =

 ∂ p S
H is a homegenous and continuous function so that the riule of Schwartz can be applied leading to
the desired expression:
  ∂ H   
  ∂ H   
 ∂  
 ∂ 
  
 
 ∂ T    ∂ S p  
∂ 2H
∂ 2 H    ∂ p  S    ∂ V 

 =
 = ∂ S ∂ p = ∂ p∂ S = 
 =∂S 
S
∂p
∂
p
 ∂ p S 


 





p

S
c)
∂ p ∂S 

 =

 ∂ T V  ∂ V T
From exercise 1 it is known that the partial derivative of A with respect to T or Vequal to p or S
are:
dA = − pdV − SdT
and
∂ A
∂ A
A = A (V , T ) ⇒ dA = 
dT
 dV + 
∂ V T
∂ T V


with this we get
∂ A
−p = 

 ∂ V T
&
∂ A
−S =

 ∂ T V
The Helmholtz free energy A is a homgeneous and contiuous function so that the rule of Schwartz
can be applied leading to the desired expression:

∂
∂ p

−
 =
 ∂ T V 



 ∂ A   
∂

 
2
2
  ∂ V T   = ∂ A = ∂ A = 

∂T
∂ V ∂ T ∂ T∂ V 


V

 ∂ A 


  ∂ T V
∂V


  = − ∂ S 



 ∂ T V

T
9) Show that:
a)

 ∂V  
dH =  V − T 
  dp + CP dT

 ∂ T  p 

In this exercise it can be seen how useful Maxwell relations are. For the derivation we start with
the total differential of the enthalpy in terms of T and p:
∂ H 
∂ H 
∂ H 
H = H ( p, T ) ⇒ dH = 
 dp = C p dT + 
 dp
 dT + 
 ∂ T p
 ∂ p T
 ∂ p T
To get the desired expression another manner to describe the partial derivative  ∂ H  needs to be


 ∂ p T
found.
From the earlier exercises we know that the total differential can also be described based on the
definition of the enthalpy leading to: dH = TdS + Vdp
Dividing this expression by dp while keeping T constant gives another description of the partial
derivative
.
∂H 
∂S 

 =T 
 +V
p
∂

T
 ∂ p T
However, this is not yet the expression we are looking for. In the next step the partial derivative
 ∂ S  needs to be rewritten. In general, partial derivatives of including the entropy can be


 ∂ p T
∂S 
 ∂V 

 = −

p
∂
 ∂ T p

T
rewritten with the help of a Maxwell relation. From exercise 2 it is known that:
Incorporating this into the expression for the partial derivative  ∂ H  gives:


 ∂ p T
∂ H 
 ∂V 

 = −T 
 +V
 ∂T p
 ∂ p T
Leading to the expression of dH:
∂ H 
∂ H 
dH = 
dT
 dp + 
∂
∂ T  p
p


T

 ∂V  
= V − T 
  dp + CP dT

 ∂ T  p 

This equation describes the change of the enthalpy for systems consisting of an arbitrary
component. The expression shows also how the change in the enthalpy can be determined
experimentally.
b)
 ∂ p

dU =  T 
 − p  dV + CV dT
  ∂ T V

The approach to show that the above expression is valid, is the same as in exercise a. We start with
the description of the internal energy U in terms of V and T and the total differential.
 ∂U 
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 dT + 
 dV = CV dT + 
 dV
 ∂ T V
 ∂ V T
 ∂ V T
To show that the above expression is valid, it needs to be shown that  ∂ U  = T ⋅  ∂p  − p




It is known that: dU = TdS − pdV
Dividing the expression by dV keeping T constant gives:
 ∂ V T
 ∂T V
 ∂U 
∂S 

 =T 
 −p
 ∂ V T
 ∂ V T
Next the partial derivative  ∂ S  needs to be rewritten. Again for this purpose an Maxwell relation


 ∂ V T
is used (see exercise 2):
∂S  ∂ p

 =

 ∂ V T  ∂ T V
Incorporating this into the partial derivative of the internal energy gives:  ∂ U  = T  ∂ p  − p
 ∂ V T
 ∂ T V
Leading to the total differential dU:
 ∂U 
 ∂U 
dU = 
 dV +  ∂ T  dT
∂
V

T

V
 ∂ p

= T 
 − p  dV + CV dT
  ∂ T V

This expression contains only properties which can be determined experimentally and is valid for
every arbitrary system consisting of a single component.
10. Show that the following expressions are valid:
 ∂U 
 = CV
 ∂T V
a.
In general: 
b.
In general:
c.
For an ideal gas:
 ∂H 

 = Cp
 ∂T  p
C p = CV + n ⋅ R
Solution:
a)
To prove that the expression
 ∂U 

 = CV is valid, an isochoric process (dV=0;V= const) needs
 ∂T V
to be described. The subscript V means that the volume is kept constant. For the description of the
change of the internal energy during an isochoric process, one starts with the first law of
thermodynmaics, followed by the description of the work:
dU = dq + dw
dw = − pex ⋅ dV
For an isochoric process (dV = 0) the work done is computed by: dw = 0 ⇒ w = 0
With this the first law of thermodynamics can be rewritten: dU = dq + dw ⇒ dU = dq
Furhter it is known that fora n isochoric process dq is described by:
Incorporating this into the first law gives
dqV = cV ⋅ dT .
dUV = dqV = cV ⋅ dT . The change of the internal
energy of the isochoric process can also be described with the help of the total differential.
Therefore, we start with the definition of the internal energy U in terms of V and T and then
forming the totl differential:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 ⋅ dV + 
 ⋅ dT . For an
 ∂V T
 ∂T V
isochoric process (dV = 0), the total differential simplifies to
 ∂U 
 ∂U 
 ∂U 
dU = 
 ⋅ dV + 
 ⋅ dT = 
 ⋅ dT . Incorporating the total differential into
 ∂V T
 ∂T V
 ∂T V
the first law, gives the desired expression:
dUV = cV ⋅ dT


 ∂U 
 ∂U 
⋅ dT ⇒ cV = 
 ⇒ cV ⋅ dT = 
 ∂U 


dUV = 
 ∂T V
 ∂T V
 ⋅ dT 
T
∂

V

For the derivation of the expression there are no further assumptions made except of that the
process is isochoric and thus the system is incompressible. Thus, this equation holds for all
systems undergoing an isochoric process.
b)
 ∂H 
 = C p an isobaric process (dp=0; p = const)
 ∂T  p
For the prove of the expression 
needs to be described. The subscript p indicates that the process is at constant pressure.
For the derivation, one starts with the first law of thermodynamics.
dU = dq + dw
dw = − pex ⋅ dV
In a next step it is assumed that the process is reversible so that pex = p and so that the work can be
computed:
dw = − p ⋅ dV
Because the process is isobaric, we switch from the internal energy to the enthalpy. Thus, first the
relation between the enthalpy and the internal energy and its relationship to the first law needs to
be established:
H ≡ U + p ⋅V ⇒ U = H − p ⋅V
⇒ dU = d ( H − p ⋅V ) = dH − d ( p ⋅ V ) = dH − p ⋅ dV − V ⋅ dp
⇒ dH − p ⋅ dV − V ⋅ dp = dq − p ⋅ dV
So, we see that fora n isobaric and reversible process:
process the heat can be computed by:
dH = dq .Additional, fora n isobaric
dq p = c p ⋅ dT so that dH p = dq p = cP ⋅ dT . The change
of the enthalpy can also be described with the help of the total differential. First, the enthalpy H is
defined in terms of p and T before its total differential is formed:
 ∂H 
 ∂H 
H = H ( p, T ) ⇒ dH = 
 ⋅ dp + 
 ⋅ dT . For an isobaric process (dp=0) this
 ∂T  p
 ∂p T
simplifies to
 ∂H 
 ∂H 
 ∂H 
dH p = 
 ⋅ dp + 
 ⋅ dT = 
 ⋅ dT .
 ∂T  p
 ∂T  p
 ∂p T
This gives the desired expression:
dH p = c p ⋅ dT


 ∂H 
 ∂H 
⋅ dT ⇒ c p = 
 ⇒ cP ⋅ dT = 
 ∂H 


dH p = 
 ∂T  p
 ∂T  p
 ⋅ dT 
 ∂T  p

For the derivation it was assumed that the process is isobaric but also reversible. There are no
further assumptions made with respect to the nature of the system. So, it can be said that this
equation is general valid for all reversible and isobaric processes!
c)
In this exercise, we need to show that: C p = CV + n ⋅ R
From the other exercises we saw that CP and CV are related to the description of either an isobaric or
an isochoric process and that these parameters are directly related to either the enthalpy or the internal
energy. Thus in order to derive this expression, we need to find a way how the change in enthalpy at
constant pressure and in internal energy at constant volume can be coupled. The total differentials of
the internal energy and of the enthalpy are given below:
 ∂U 
 ∂U 
dU = 
 ⋅ dV + 
 ⋅ dT
 ∂V T
 ∂T V
 ∂H 
 ∂H 
dH = 
 ⋅ dp + 
 ⋅ dT
 ∂T  p
 ∂p T
We know that:
 ∂U 
CV = 

 ∂T V
 ∂H 
CP = 

 ∂T  p
Further, we know that the internal energy U and the enthalpie H of an ideal gas only change due to
changes in temperature but not due to changes of the pressure or volume. A change in pressure or
volume would mean that the distance between the gas molecules is changed. However, for an ideal gas
the molecules do not have interactions with each other so that a change of the distance between the
molecules does not have any effect. Thus for an ideal gas undergoing an isothermal process the
following expressions hold:
 ∂U 

 =0
 ∂V T
 ∂H 

 =0
 ∂p T
With this we get for the total differential describing the change of the internal energy or the enthalpy of
ani deal gas:
dE = CV ⋅ dT
dH = C p ⋅ dT
These equations hold for ideal gases undergoing an isothermal, an isobaric, an isochoric and an
adiabatic process.
To show the validity of the above given expression, the change of internal energy and enthalpy is
couple via the definition of the enthalpy and the subsequent total differential:
H ≡ U + p ⋅V ⇒ dH = d (U + p ⋅ V ) = dU + d ( p ⋅ V )
As a last step, the product ‘pv’ is replaced by the ideal gas law, leading to the desired equation:
dH = dU + d ( p ⋅ V )
dH = dU + d ( n ⋅ R ⋅ T ) 

dH = cP ⋅ dT
 ⇒ cP ⋅ dT = cV ⋅ dT + n ⋅ R ⋅ dT = ( cV + n ⋅ R ) ⋅ dT

dU = cV ⋅ dT

⇒ cP = cV + n ⋅ R
However, this expression is not generally valid. It only holds for ideal gases.
11. 5 mol of an ideal gas expands isobaric and reversible from its initial volume V1 to its final
volume V2. The final volume is three times the initial volume.
The initial state is described by the following values:
V1 = 0.073 m3 en T1 = 350 K,
The final state by: V2 = 3·V1.
Determine using the given data for the process: ∆U, ∆H, q en w.
Further the specific heat capacity of the ideal gas is given:
C p = 36,304
J
= cons tan t
mol ⋅ K
Solution:
The system is closed (dn=0; n = const) and consists of 5 mol ideal gas. For an ideal gas, the ideal gas law is
valid: p ⋅ V = n ⋅ R ⋅ T . Additionally, the change of the internal energy and of the enthalpy at constant
temperature is zero for an ideal gas:
 ∂E 

 =0
 ∂V T
 ∂H 

 =0
 ∂p T
The process is reversible (pex = psysteem = p) and isobaric (dp=0; p = const). For the initial state the following
conditions are given:
V1 = 0.073m3
T1 = 350 K ⇒ p1 = p =
n ⋅ R ⋅ T1
V1
For the final state we know that: V2 = 3 ⋅ V1 = 3 ⋅ 0.073m
p2 = p1 = p
The temperature at the final state is not given but can be computed using the ideal gas law taking into
account that the pressure is kept constant during the process:
3
p2 =
n ⋅ R ⋅ T2
n ⋅ R ⋅ T1
V
3 ⋅V1
= p1 =
⇒ T2 = 2 ⋅ T1 =
⋅ T1 = 3 ⋅ T1 = 3 ⋅ 350 K
V2
V1
V1
V1
⇒ T2 = 1050 K
Now we have all the data we need to do the remaining computations. As before we start with the first law:
dU = dq + dw
V2
dw = − pex ⋅ dV = − p ⋅ dV
reversible
process
⇒ w = ∫ dw = − ∫ p ⋅ dV
V1
V2
⇒ w = − p ⋅ ∫ dV = − p ⋅ (V2 − V1 ) = −
isobaric
V1
⇒w=−
n ⋅ R ⋅ T1
⋅ ( 3 ⋅V1 − V1 )
V1
n ⋅ R ⋅ T1
J
⋅V1 ( 3 − 1) = −n ⋅ R ⋅ T1 ⋅ 2 = −5mol ⋅ 8.314
⋅ 350 K ⋅ 2
V1
mol ⋅ K
⇒ w = −29099 J
As a next step the change of enthalpy is computed (we do this before the computation of the
change of the internal energy because we deal with an isobaric process). We describe the enthalpy
H in terms of T and p: H(p,T). The change of the enthalpy can be described by the total
differential:
 ∂H 
 ∂H 
dH = 
 ⋅ dp + 
 ⋅ dT and for an isobaric process (dp=0) we get:
 ∂T  p
 ∂p T
 ∂H 
 ∂H 
dH = 
 ⋅ dT and because of 
 = C p we get: dH = C p ⋅ dT . The molar heat capacity is
 ∂T  p
 ∂T  p
given by:
C p = n ⋅ C p , so that the change in the enthalpy can be computed by:
T2
T2
T1
T1
∆H = ∫ dH = ∫ n ⋅ C p ⋅ dT = n ⋅ C p ⋅ ∫ dT
assu min g C p = const
∆H = n ⋅ C p ⋅ (T2 − T1 ) = n ⋅ C p ⋅ ( 3 ⋅ T1 − T1 ) = 5mol ⋅ 36.304
∆H = 127064 J = 127.1kJ
J
⋅ 2 ⋅ 350 K
mol ⋅ K
With this the change of the internal energy can be computed:
H ≡ U + p ⋅ V ⇒ U = H − p ⋅ V ⇒ dU = dH − d ( p ⋅V )
ideal
gas
p ⋅V = n ⋅ R ⋅ T ⇒ dU = dH − d ( n ⋅ R ⋅ T ) = dH − n ⋅ R ⋅ dT
T2
∆U = ∫ dU = ∫ dH − n ⋅ R ⋅ ∫ dT = ∆H − n ⋅ R ⋅ (T2 − T1 ) = ∆H − n ⋅ R ⋅ ( 3 ⋅ T1 − T1 ) = ∆H − n ⋅ R ⋅ 2 ⋅ T1
T1
∆U = 127064 J − 5mol ⋅ 8.314
∆U = 97965 J = 97.97kJ
J
⋅ 2 ⋅ 350 K
mol ⋅ K
The heat exchanged during the process can be computed with the first law:
dU = dq + dw ⇒ dq = dU − dw ⇒ q = ∆U − w = 97965 J − ( −29099 J )
⇒ q = 127064 J = 127.064kJ
The heat exchanged during the process can also be computed by:
dH = cP ⋅ dT = d (U + p ⋅ V ) = dq + dw + p ⋅ dV + V ⋅ dp = dq − p ⋅ dV + p ⋅ dV + V ⋅ dp
isobaric ⇒ dH = cP ⋅ dT = dq ⇒ ∆H = q = 127064 J = 127.064kJ
12. 4 mol of an ideal gas undergoes a reversible process from its initial state at 250C and 1 bar to
its final state of 900C and 10 bar. To reach the final state, first the ideal gas is heated at
constant volume from 25 oC to 90 oC. In a next step the gas is isothermally compressed to a
pressure of 10 bar. Please compute q, w, ∆E en ∆H of the complete process. Assume for the
CP =
computation that:
7
⋅R.
2
Solution:
The system is closed (dn=0; n = const) and consists of 4 mol ideal gas. Because the system consists of an
ideal gas the ideal gas law can be applied: p ⋅ V = n ⋅ R ⋅ T . The change of the internal energy and of the
enthalpy at constant temperature for an ideal gas is zero, meaning there is no change. This is described
mathematically by:
 ∂U 

 =0
 ∂V T
 ∂H 

 =0
 ∂p T
The process is reversible so that pex = psysteem = p and split into two subprocesses.
Step1: reversible and isochoric (dV= 0; V = const);
Initial state: T1= 250C = 298.15 K and p1= 1bar = 105 N m-2
End state: T2 = 90 0C = 363.15 K; V2 = V1
Step 2: reversible and isothermal (dT = 0; T = const);
Initial state= end state of step 1: T2 = 90 0C = 363.15 K; V2 = V1
Eind state = end state of the total process: T3 = 90 0C = 363.15 K;
p3 = 10 bar = 10·105 N m-2
Description of step 1:
We start with the first law followed by the computation of the work….:
dU = dq + dw
dw = − pex ⋅ dV = − p ⋅ dV
V2
reversible
⇒ w = ∫ dw = − ∫ p ⋅ dV
V1
isochoric
dV = 0 ⇒ wstep1 = 0 J
For the computation of the change of the internal energy we start with the total differential:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 ⋅ dV + 
 ⋅ dT
 ∂V T
 ∂T V


isochoric ⇒ dV = 0


 ∂U 
=
ideal gas ⇒ 
0
 ⇒ dU = CV ⋅ dT

 ∂V T


 ∂U 
general CV = 


 ∂T V 
There is no value given for CV however for an ideal gas we know that: C p = CV + n ⋅ R so that
CV = CP − n ⋅ R ⇒ n ⋅ CV = n ⋅ CP − n ⋅ R ⇒ CV = CP − R and thus
(
)
dU = n ⋅ CV ⋅ dT = n ⋅ CP − R ⋅ dT
T2
(
)
(
)
T2
(
)
⇒ ∆U = ∫ dU = ∫ n ⋅ CP − R ⋅ dT = n ⋅ CP − R ⋅ ∫ dT = n ⋅ CP − R ⋅ (T2 − T1 )
T1
T1
5
J
7

⇒ ∆U = 4mol ⋅  ⋅ R − R  ⋅ ( 363.15 − 298.15 ) K = 4mol ⋅ ⋅ 8.314
⋅ 65 K
2
mol ⋅ K
2

⇒ ∆U step1 = 5404.1J
The heat exchanged can be computed with the help of the first law:
dU = dq + dw ⇒ dq = dU − dw ⇒ q = ∆U − w = 5404.1J − 0 J
⇒ qstep1 = 5404.1J
Next, the change of the enthalpy during the process is computed. This can be done in different ways. Here
both ways are shown.
Manner 1: The enthalpy H is a function of T and p: H(p,T). The change of the enthalpy is described by the
total differential:
 ∂H 
 ∂H 
 ∂H 
dH = 
 ⋅ dp + 
 ⋅ dT with 
 = C p and because the system
 ∂T  p
 ∂T  p
 ∂p T
consists of an ideal gas the first derivative of the enthalpy with respect to the pressure at constant pressure
 ∂H 

 = 0 so that the total differential of the enthalpy simplifies to: dH = C p ⋅ dT .
 ∂p T
With the given molar heat capacity C p = n ⋅ C p the change of the enthalpy can be computed:
is equal to zero:
T2
T2
T1
T1
∆H = ∫ dH = ∫ n ⋅ C p ⋅ dT = n ⋅ C p ⋅ ∫ dT
because C p = const
7
7
J
∆H = n ⋅ C p ⋅ (T2 − T1 ) = 4mol ⋅ ⋅ R ⋅ 65 K = 4mol ⋅ ⋅ 8.314
⋅ 65 K
2
2
mol ⋅ K
∆H step1 = 7565.74 J
Manner 2:
The other way to compute the change of the enthalpy is via the definition of the enthalpy and the ideal gas
law.
H ≡ U + p ⋅ V ⇒ dH = dU + d ( p ⋅ V ) = dU + p ⋅ dV + V ⋅ dp
isochoric ⇒ dH = dU + V ⋅ dp
p2
⇒ ∆H = ∫ dH = ∫ dU + ∫ V ⋅ dp
p1
p2
isochoric
V2 = V1 ⇒ ∆H = ∆U + V1 ⋅ ∫ dp
p1
p

n ⋅ R ⋅ T1
⋅ ( p2 − p1 ) = ∆U + n ⋅ R ⋅ T1 ⋅  2 − 1
p1
 p1 
n ⋅ R ⋅ T2 n ⋅ R ⋅ T2 n ⋅ R ⋅ T2 p1 ⋅ T2
p2 =
=
=
=
n ⋅ R ⋅ T1
V2
V1
T1
p1
⇒ ∆H = ∆U + V1 ⋅ ( p2 − p1 ) = ∆U +
 p1 ⋅ T2

 T

T

⇒ ∆H = ∆U + n ⋅ R ⋅ T1 ⋅  1 − 1 = ∆U + n ⋅ R ⋅ T1 ⋅  2 − 1 = ∆U + n ⋅ R ⋅ (T2 − T1 )
 p1

 T1 




J
⇒ ∆H = 5404.1J + 4mol ⋅ 8.314
⋅ ( 363.15 − 298.15 ) K
mol ⋅ K
⇒ ∆H step1 = 7565.74 J
Description of step 2: reversible and isothermal
We start with the first law of thermodynamics followed by the computation of the work etc:
dU = dq + dw
V3
dw = − pex ⋅ dV = − p ⋅ dV
reversible
⇒ w = ∫ dw = − ∫ p ⋅ dV
V2
For the given process, the work cannot directly be computed because the pressure and the volume
change during the process. In order to allow the computation of the work, the pressure is replaced by the
ideal gas law:
3
n ⋅ R ⋅T
n ⋅ R ⋅T
⇒ w = −∫
⋅ dV
V
V
V2
V
p=
V 
dV
= −n ⋅ R ⋅ T ⋅ ln  3 
V
 V2 
V2
V3
isothermal ⇒ w = −n ⋅ R ⋅ T ⋅ ∫
The volumes of the inital and the end state, V2 and V3, are given if not directly. The volumes can be
computed with the given data and the ideal gas law. For the computation of the initial state of step 2 (V2)
one needs to remember that the first process step is isochoric (V1 = V2). The volume V1 is computed with
the ideal gas law and with the given data:
V=
n ⋅ R ⋅T
p
n ⋅ R ⋅ T1 
n ⋅ R ⋅ T3

p1  V3
p3
T ⋅p
=
= 3 1
⇒
n ⋅ R ⋅ T3
 V2 n ⋅ R ⋅ T1 T1 ⋅ p3
V3 =
p1

p3
T ⋅ p 
J
 363.15 K ⋅1bar 
⇒ w = −n ⋅ R ⋅ T ⋅ ln  3 1  = −4mol ⋅ 8.314
⋅ 363.15 K ⋅ ln 

mol ⋅ K
 298.15 K ⋅10bar 
 T1 ⋅ p3 
V2 = V1 =
⇒ wstap 2 = 25426.3 J
Next the change of the internal energy is computed. For this purpose we start with the description of the
change of the internal energy by the total differential:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 ⋅ dV + 
 ⋅ dT
 ∂V T
 ∂T V


isothermal ⇒ dT = 0


 ∂U 
ideal gas ⇒ 
=
0
 ⇒ dU = 0 ⇒ ∆U step 2 = 0 J

 ∂V T


 ∂U 
general CV = 


 ∂T V 
The exchanged heat is computed with the help of the first law of thermodynamics:
dU = dq + dw ⇒ dq = dU − dw ⇒ q = ∆U − w = − w
⇒ qstep 2 = − wstep 2 = −25426.3J
De verandering van de enthalpie kan weer op verschillende manieren worden berekend. Ik geef een
manier als voorbeeld:
H ≡ E + p ⋅ V ⇒ dH = dE + d ( p ⋅ V )
ideaal
gaswet
p ⋅V = n ⋅ R ⋅ T ⇒ dH = dE + d ( n ⋅ R ⋅ T ) = dE + n ⋅ R ⋅ dT
isotherm ⇒ dH = 0 + 0 ⇒ ∆H stap 2 = 0 J
For the changes of the complete process, the changes of the subprocesses need tob e summed up:
∆U = ∆U step1 + ∆U step 2 = 5404.1J + 0 J = 5404.1J
∆H = ∆H step1 + ∆H step 2 = 7565.74 J + 0 J = 7565.74 J
q = qstep1 + qstep 2 = 5404.1J − 25426.3J = −20022.2 J
w = wstep1 + wstep 2 = 0 J + 25426.3 J = 25426.3 J
13. 3 mol of an ideal gas suddenly expands at a constant temperature of 350 K against a constant
external pressure of 1 bar. Initially the ideal gas occupies a volume of 10 L, in the end a
volume of 15 L. Determine with the help of the given data ∆U, ∆H, q en w of the gas
undergoing the described process.
The system is closed (dn=0; n = const) and consists of 3 mol ideal gas. The system consists of an ideal gas
so that the ideal gas law can be applied: p ⋅ V = n ⋅ R ⋅ T . Further it is known that the internal energy U
and the enthalpy H for a system consisting of an ideal gas only changes with temperature. Thus, we can
write:
 ∂U 

 =0
 ∂V T
 ∂H 

 =0
 ∂p T
Because the process is spontaneous, the process is irreversible. The external pressure is constant and equal
to: pex = 1 bar.
Further, the process is isothermal (dT = 0; T = const = 350 K);
Initial state: V1 = 10 L = 10·10-3 m3
End state : V2 = 15 L = 15·10-3 m3
With the help of the ideal gas law, the pressure at the initia land the end state can be computed.
For the thermodynamic description of the process one starts with the first law of thermodynamics, followed
by the computation of the work etc….:
dU = dq + dw
V3
dw = − pex ⋅ dV ⇒ w = ∫ dw = − pex ⋅ ∫ dV = − pex ⋅ (V2 − V1 )
V2
⇒ w = −1bar ⋅ (15 − 10 ) L = −1 ⋅105
N
⋅ 15 − 10 ) ⋅10−3 m3
2 (
m
⇒ w = −5 ⋅102 Nm = −500 J
Next the change of the internal energy can be computed starting with the total differential of the internal
energy:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 ⋅ dV + 
 ⋅ dT
 ∂V T
 ∂T V


isothermal ⇒ dT = 0


 ∂E 
ideal gas ⇒ 
 = 0  ⇒ dU = 0 ⇒ ∆U = 0 J
 ∂V T


 ∂E 
general CV = 


 ∂T V 
The heat exchanged during the process is computed with the first law of thermodynamics:
dU = dq + dw ⇒ dq = dU − dw ⇒ q = ∆U − w = − w
⇒ q = − w = 500 J
The change of the enthalpy can be computed in different ways. I only give one manner, the other one can
be seen in exercise 3 step 1.:
The enthalpy H is a function of T and p: H(p,T). The change of the enthalpy can be computed
 ∂H 
 ∂H 
 ∂H 
 ⋅ dp + 
 ⋅ dT with 
 = Cp .
 ∂T  p
 ∂T  p
 ∂p T
based on the total differential of the enthalpy: dH = 
 ∂H 
 = 0 so that: dH = C p ⋅ dT .
 ∂p T
Because the system consists of an ideal gas, we can further write: 
The process is isothermal (dT=0) so that:
T2
∆H = ∫ dH = ∫ C p ⋅ dT = 0 J
T1
14. 2 mol of an ideal gas is cooled reversible and adiabatic from 298 K to 260 K. During this
process the volume changes from V1 to V2 = 3·V1. For the computations you can assume that
the heat capacities, CV en CP, are constant and do not depend on the temperature.
a. Compute, using the given data, the heat capacity at constant volume, CV [J K-1] , of
the gas.
C p [J K-1 mol-1].
b.
Then compute the heat capacity at constant pressure,
c.
Finally compute for the process ∆U, ∆H, q en w using the computed values for the
heat capacities at constant volume and constant pressure.
Solution:
The system is closed (dn=0; n = const) consists of 2 mol ideal gas. Further the ideal gas is valid because the
system consists of an ideal gas: p ⋅ V = n ⋅ R ⋅ T . For a system consisting of an ideal ga, the internal
energy U and the enthalpy H only change with temperature so that we can write:
 ∂U 

 =0
 ∂V T
 ∂H 

 =0
 ∂p T
Further, fora n ideal gas CP and CV coupled by (see exercise 1c): CP= CV + n·R
The process is reversible (pex = psysteem = p) and adiabatic(dq=0; q = 0 J). Further the initial state is
described by:
V1
T1 = 298 K
The end state by: V2 = 3 ⋅ V1
T2 = 260 K
To compute CV the process needs tob e described thermodynamically. Therefore, we start with the first law
of thermodynamics and with the computation of the work:
dU = dq + dw
dw = − pex ⋅ dV = − p ⋅ dV = −
n ⋅ R ⋅T
⋅ dV
V
reversible
Note, that during an adiabatic process the temperature T, the pressure p and the volume V change so that
the equation for the work cannot be integrated!
According to the first law we get:
dU = dq + dw
adiabatic ⇒ dq = 0 ⇒ dE = dw
Thus the computation of the work can be determined via the change of the internal energy . The change of
the internal energy is described by the total differential:
 ∂U 
 ∂U 
dU = 
 ⋅ dV + 
 ⋅ dT and with
 ∂V T
 ∂T V
dU = CV ⋅ dT
And finally
 ∂U 

 = CV and for an ideal gas
 ∂T V
 ∂U 

 = 0 we get:
 ∂V T
dU = dq + dw = dw
CV ⋅ dT = − p ⋅ dV
CV ⋅ dT = −
n ⋅ R ⋅T
⋅ dV
V
The last equation can be solved by splitting of the variables:
2
2
n ⋅ R ⋅T
dT
dV
dT
dV
CV ⋅ dT = −
⋅ dV ⇒ CV ⋅
= −n ⋅ R ⋅
⇒ ∫ CV ⋅
= −∫ n⋅ R⋅
V
T
V
T
V
T1
V1
T
T2
⇒ CV ⋅ ∫
T1
V
2
T 
V 
dT
dV
= −n ⋅ R ⋅ ∫
⇒ CV ⋅ ln  2  = − n ⋅ R ⋅ ln  2 
T
V
 T1 
 V1 
V1
V
V 
ln  2 
V
⇒ CV = −n ⋅ R ⋅  1 
T 
ln  2 
 T1 
With this equation and the given parameters, CV can be computed:
V 
 3 ⋅V1 
ln  2 
ln 

ln ( 3)
V1 
V1 
J
J


= −2mol ⋅ 8.314
⋅
= −2mol ⋅ 8.314
⋅
CV = −n ⋅ R ⋅
mol ⋅ K
mol ⋅ K
T 
 260 K 
 260 K 
ln 
ln 
ln  2 


 298 K 
 298 K 
 T1 
CV = 133.916
J
K
b) Now, the heat capacity at constant pressure needs to be computed. For this purpose we make use of the
following equation which is valid for an ideal gas: CP= CV + n·R
The molar heat capacity is given by:
C p = n ⋅ C p , so that we can computed the actual value for the
C p = CV + n ⋅ R ⇒ C p = CV + R ⇒ C p =
heat capacity at constant pressure
Cp =
c)
133.916
2mol
J
K + 8.314
CV
+R
n
J
J
= 75.27
mol ⋅ K
mol ⋅ K
Now all the necessary data are computed to allow the computation of ∆U, ∆H, q en w:
The process is adiabatic so that: dq = 0 and q = 0 J.
The change of the internal energy can be computed with the equation given in exercise a:
dU = dw
T2
T2
T1
T1
dU = CV ⋅ dT ⇒ ∆U = ∫ dU = ∫ CV ⋅ dT = CV ⋅ ∫ dT = CV ⋅ (T2 − T1 )
∆U = w = 133.916
J
⋅ ( 260 − 298 ) K = 5088.81J
K
For the computation of ∆H we recall that of each process an ideal gas undergoes, the change of
enthalpy can be computed by:
T2
T2
T1
T1
∆H = ∫ dH = ∫ C p ⋅ dT = C p ⋅ ∫ dT = n ⋅ C p ⋅ (T2 − T1 )
∆H = 2mol ⋅ 75.27
J
⋅ ( 260 − 298 ) K = −5720.52 J
mol ⋅ K
15. 8 mol of an ideal gas expands from its initial volume V1 to its final volume V2.
The initial state is described by:
V1 = 0.15 m3 en p1 = 1.5 bar,
The final state by: V2 = 0.22 m3.
Compute with the given data ∆U, ∆H, q en w for the following processes:
a.
b.
c.
isothermal and reversible.
Isothermal at a constant external pressure of pex = 1bar.
Explain, using also the computed values of exercises a and b which of these
parameters (∆U, ∆H, q en w) are state functions.
Solution:
The system is closed (dn=0; n = const) and consists of 8 mol ideal gas so that the ideal gas law is valid:
p ⋅ V = n ⋅ R ⋅ T For a system consisting of an ideal gas there is no change of the internal energy and
enthalpy for an isothermal process:
 ∂U 

 =0
 ∂V T
 ∂H 

 =0
 ∂p T
The process to be described is an isothermal expansion (dT = 0; T = const).
a)
In the first exercise the change of the internal energy, the enthalpy and the exchanged heat and
work for the reversible process is wanted.
The fact that the process is done reversibly means that during the complete process the external
pressure pex is equal to the pressure of the system (pex = psystem = p).
For the thermodynamic process description we start with the first law of thermodynamics followed by
the computation of the work and heat:
dU = dq + dw
V3
dw = − pex ⋅ dV = − p ⋅ dV ⇒ w = ∫ dw = − ∫ p ⋅ dV
V2
The process is neither isobaric nor isochoric so that the equation to calculate the work cannot be
integrated (because V and p vary during the process). However, as we deal with an ideal gas the ideal
gas law is incorporated replacing the pressure in terms of volume and temperature. In this way, the
term is rewritten in terms of V and T and because it is an isothermal process the equation can be
integrated (only the volume varies):
2
n ⋅ R ⋅T
n ⋅ R ⋅T
n ⋅ R ⋅T
p=
⇒ dw = − p ⋅ dV = −
⋅ dV ⇒ w = ∫ dw = − ∫
⋅ dV
V
V
V
V1
V
V2
⇒ w = −n ⋅ R ⋅ T ⋅ ∫
V1
V 
dV
= −n ⋅ R ⋅ T ⋅ ln  2 
V
 V1 
In order to actual compute the work, the temperature needs to be computed with the given data and the
ideal gas law:
p ⋅V
p ⋅V
p ⋅V
⇒ T1 = 1 1 = T2 = 2 2 = T
n⋅R
n⋅R
n⋅R
V 
V 
p ⋅V
⇒ w = −n ⋅ R ⋅ 1 1 ⋅ ln  2  = − p1 ⋅V1 ⋅ ln  2 
n⋅R
 V1 
 V1 
T=
 0.22m3 
 0.22m3 
5 N
3
1.5
10
0.15
ln
⇒ w = −1.5bar ⋅ 0.15m3 ⋅ ln 
=
−
⋅
⋅
m
⋅

3 
3 
m2
 0.15m 
 0.15m 
⇒ w = −8617.33 J
Next the change of the internal energy is computed:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 ⋅ dV + 
 ⋅ dT
 ∂V T
 ∂T V


isothermal ⇒ dT = 0


 ∂U 
ideal gas ⇒ 
 = 0  ⇒ dU = 0 ⇒ ∆U = 0 J
 ∂V T


∂
U


general CV = 


 ∂T V 
With the computed work and the computed change of the internal energy the heat can be computed
using the first law of thermodynamics:
dU = dq + dw ⇒ dq = dU − dw ⇒ q = ∆U − w = − w
⇒ q = − w = 8617.33 J
The change of the enthalpy can be computed in different manners. Only one way is given, the others
have been given in other exercises:
The enthalpy is a function of T and p: H(p,T). The change of the enthalpy can be determined with
 ∂H 
 ∂H 
 ∂H 
dH = 
 ⋅ dp + 
 ⋅ dT with 
 = C p and
 ∂T  p
 ∂T  p
 ∂p T
 ∂H 
because the system consists of an ideal gas: 
 = 0 so that: dH = C p ⋅ dT . The process is
 ∂p T
the help of the total differential:
isothermal (dT=0) so that:
T2
∆H = ∫ dH = ∫ C p ⋅ dT = 0 J
T1
b) For this exercise the changes in the internal energy ∆U, the enthalpy ∆H, the exchanged heat q
and work w are asked if the ideal gas undergoes an isothermal process against a constant external
pressure of 1 bar.
The expression ‘against a constant external pressure’ means that the process is irreversible.
The approach to compute the asked properties is the same as in exercise a:
dU = dq + dw
dw = − pex ⋅ dV
The work is now calculated without replacing the external pressure by the pressure of the system
because the process is not reversible. Thus
V2
V2
V1
V1
dw = − pex ⋅ dV ⇒ w = ∫ dw = − ∫ pex ⋅ dV = − pex ⋅ ∫ dV
⇒ w = − pex ⋅ (V2 − V1 ) = −1bar ⋅ ( 0.22 − 0.15 ) m3 = −1⋅105
N
⋅ ( 0.22 − 0.15 ) m3
m2
⇒ w = −7000 J
There is no difference for the change of the internal energy compared to exercise a. For the
description the total differential is used and all the conditions for an ideal are incorporated so that:
 ∂U 
 ∂U 
U = U (V , T ) ⇒ dU = 
 ⋅ dV + 
 ⋅ dT
 ∂V T
 ∂T V


isothermal ⇒ dT = 0


 ∂U 
ideal gas ⇒ 
=
0
 ⇒ dU = 0 ⇒ ∆U = 0 J

 ∂V T


 ∂U 
general CV = 


 ∂T V 
The heat is computed using the first law of thermodynamics:
dU = dq + dw ⇒ dq = dU − dw ⇒ q = ∆U − w = − w
⇒ q = − w = 7000 J
The change of enthalpy during the process can be computed in different manners. Again only one
of the possible ways to compute the change of the enthalpy is given, namely via the total
differential:
 ∂H 
 ∂H 
 ∂H 
dH = 
 ⋅ dp + 
 = C p Additionally, for an ideal
 ⋅ dT with 
 ∂T  p
 ∂T  p
 ∂p T
 ∂H 
 = 0 So we get: dH = C p ⋅ dT . The process is isothermal (dT=0):
 ∂p T
gas: 
T2
∆H = ∫ dH = ∫ C p ⋅ dT = 0 J
T1
d.
Show with the help of the calculated properties of exercise a and b which of the
properties are state functions and which not.
Process exercise a
Process exercise b
0
0
∆U [J]
∆H [J]
0
0
q
w
[J]
[J]
8617,33
-8617,33
7000
-7000
The change of a state function does not change for a process as long as the initial and end state are
the same. From this we can see that the internal energy and the enthalpy are state functions while
the heat and work are no state functions because they depend on the way the process is done. Thus
the values of the exchanged heat and work for the two processes differ.