Behavior of Various Thermodynamic Properties for Several Processes
Property
Symbol
Units
Water
Melting
Heating
Ideal Gas
Boiling
Isothermal
Isobaricb
Isochoricb
Volume
V
m3
ΔV =  Vo ΔT (smalla)
ΔV is negative
ΔV is positive
Vf = Vi (Pi/Pf)
Vf = Vi (Tf/Ti)
Constant
Pressure
P
N/m2
Constant (assumed)c
Constantc
Constantc
Pf = Pi (Vi/Vf)
Constant
Pf = Pi (Tf/Ti)
Temp.
T
K
ΔT is given
T = 273.15K
(constant)
T = 373.15K
(constant)
Constant
Tf = Ti (Vf/Vi)
Tf = Ti (Pf/P i)
Internal
Energyd
Worke
ΔU
J
m Lf - P ΔV
m Lv - P ΔV
0
W
J
ΔV ≈ 0, so W ≈ 0 and
ΔU ≈ m c ΔT = n C ΔT
P ΔV = P  Vo ΔT (smalla)
P ΔV (< 0)
P ΔV (> 0)
n R T ln(Vf/Vi)
Heat
Q
J
m c ΔT or n C ΔT
m Lf
m Lv
n R T ln(Vf/Vi)
(df/2) n R ΔT;
(df/2) P ΔV
n R ΔT; P ΔV
[(df + 2)/2] n R ΔT;
[(df + 2)/2] P ΔV
(df/2) n R ΔT;
(df/2) V ΔP
0
(df/2) n R ΔT;
(df/2) V ΔP
Entropyf
ΔS
J/K
m c ln(Tf/Ti) or n C ln(Tf/Ti)
m Lf / T
m Lv / T
n R ln(Vf/Vi);
Q/T
[(df + 2)/2] n R ln(Tf/Ti)
(df/2) n R ln(Tf/Ti)
Adiabaticb
Vf = Vi (Ti/Tf)df/2
Vf = Vi (Pi/Pf)df/(df+2)
Pf = Pi (Tf/Ti)(df/2) + 1
Pf = Pi (Vi/Vf)1+(2/df)
Tf = Ti (Vi/Vf)2/df
Tf = Ti (Pf/P i)2/(df+2)
(df/2) n R ΔT
(both P & V change)
- (df/2) n R ΔT
0
0
NOTES:
a
The thermal expansion coefficient of water varies with temperature, from -68 X 10-6 K-1 at the freezing point to +750 X 10-6 K-1 at the boiling point. The resulting work is nonzero but negligible compared to the heat added.
The quantity denoted by “df” is the number of degrees of freedom of each molecule. Thus, for a monoatomic gas like He or Ar that has only three translational degrees of
freedom (no rotation or vibration possible), the value of df is 3. For diatomics like nitrogen and oxygen whose rotational degrees of freedom are excited at normal temperatures
(but vibrational ones are not), the value of df is 5. The internal energy of n moles of any gas is then U = (df/2) n R T. NOTE: Most textbooks do not express these quantities in
terms of degrees of freedom. Instead, they use the quantity γ = 1 + (2/df).
b
c
Usually when one warms, melts, or boils water, it is in a container that is open to the atmosphere. Thus, here it is assumed that this is the case and thus pressure is held constant
at 1 atmosphere (101325 N/m2). Of course, if water is warmed, melted, or boiled in a sealed container, the pressure will not remain constant.
The First Law of Thermodynamics requires that ΔU = Q – W. This is used to find ΔU for the three processes involving water. For an ideal gas, the internal energy depends
only on the temperature, so that ΔU = (df/2) n R ΔT for all ideal gas processes. This is then used to find the heat for the isochoric process (where no work is done), and to find the
work for the adiabatic (where no heat is transferred).
d
Vf
e
The work done by the system can always be found by using W =
∫ P dV , which is true whether the gas is ideal or not. If P = constant, which is the case for
Vi
all the water processes and the isobaric ideal gas process, then this reduces simply to W = P ΔV = P
necessary, as is the case with the isothermal and adiabatic processes.
f
n R ΔT
= n R ΔT. If P changes during the process, then an integral is
P
The entropy change can be found for a constant temperature process by ΔS = Q/T. If the temperature changes, then an integral is necessary: ΔS =
Isothermal: T is a constant (ΔU = 0);
P V = n R T (ideal gas);
Q = ΔU + W (true for all gases, not just ideal ones)
Isobaric: P is a constant;
Isochoric: V is a constant (W = 0);
∫
dQ
.
T
Adiabatic: Q = 0