Isothermal and Adiabatic Processes Preamble: Macroscopic systems, such as a glass of water, are specified by thermodynamics (state) variables, such as P,V,T ... that can be specified by equation of state P (V,T ) , which relates three of these state variables. For this course the most relevant equation of states are: Ideal Gas Equation: PV = NkBT = nRT , N ≡ number of particles, k B = 1.381× 10 −23 J i K −1 , n ≡ number of moles, R = 8.314J i K −1 i mol −1 f
f
Internal Energy Equation of Ideal Gas: U = NkBT = nRT 2
2
Quadratic Degree of Freedom: In the above f ≡ quadratic degree of freedom, which is usually discussed in more advanced courses. For monatomic gases (Helium, Argon,…) f = 3 . For diatomic gases ( O2 , N 2 ,…) f = 5 . f
f
Both PV = NkBT = nRT and U = NkBT = nRT are equation states, that can be 2
2
related by performing a Legendre Transformation. Examples of Thermodynamics processes Below are three thermodynamics processes: 1. Isotherm at constant temperature T1. This takes system from pressure P1, Temperature V1 to a state with P2 and V2. Here V2 < V1, so this is a compression. 2. Isotherm at higher constant temperature T2 > T1. This takes system state with P3, V2 to a state with P1 and V1. 3. Adiabatic process (no heat flow Q = 0) from state with P1, V1, T1 to state with P3, V2, T2. In this process the temperature increases from T1 to T2. P3 P Isotherm at T2 R E S S P2 U adiabat R E Isotherm at T1 P1 V2 VOLUME V1 Isothermal Process For the isothermal process at T1, assume that this involves one mole (n = 1 mol) helium (monatomic ideal gas) at T1 = 300K that compresses the gas from V1 = 10L = 0.01m 3 to V2 = 5L = 0.005m 3 . Find the heat flow into (Q > 0) or out (Q<0) of the system. f
3
Solution: Assume monatomic ideal gas PV = nRT and U = nRT = nRT where 2
2
3
3
f = . First Law ΔU = W + Q . But ΔU = nRΔT , since ΔT = 0 (isothermal). 2
2
V2 nRT
V2 dV
dV = −nRT ∫
Hence W = −Q . W = − ∫
, where we can pull nRT out of the V1
V1 V
V
V
V
V
integral since T = T1 = 300K. W = −nRT [ lnV ]V2 = −nRT ln 2 = nRT ln 1 . 1
V1
V2
V
V 10L
Q = −W = −nRT ln 1 Use 1 =
= 2 and T = T1 = 300K. V2
V2 5L
Q = − (1mol ) ( 8.314J i mol −1 i K −1 ) ( 300K ) = −1730J Since Q < 0, 1739 J of heat flows into the gas. Adiabatic Processes of Ideal Gas For the adiabat with P1, V1, T1 to state with P3, V2, T2. We have the following relations: 1. P1V1κ = P3V2κ , with κ = CP / CV , where κ = 3 / 2 for monatomic gas, and κ = 5 / 2 for diatomic gas. 2. V1T1 f /2 = V2T2f /2 , with f ≡ quadratic degree of freedom, where f = 3 for monatomic gas, and f = 5 , for diatomic gases at sufficiently low temperature (but not too low). For the adiabatic process, assume that this is one mole (n = 1 mol) helium (monatomic ideal gas) at T1 = 300K that compresses the gas from V1 = 10L = 0.01m 3
to V1 = 5L = 0.005m 3 . Find the work done W and verify that it equals ΔU f
3
Solution: Assume monatomic ideal gas PV = nRT and U = nRT = nRT where 2
2
3
f = . First Law ΔU = Q + W . For an adiabatic process, Q = 0 → W = ΔU . 2
V2
W = − ∫ P dV , but for an adiabat, P1V1κ = P3V2κ with κ = 5 / 3 , or simply PV 5/3 = P1V15/3 , V1
V
and P =
2
V2 dV
P1V15/3
5/3
2/3 ⎡ 3 1 ⎤
. T
he w
ork d
one b
ecomes W
=
−P
V
=
−P
V
V
−
1 1 ∫
1 1 1
⎢⎣ 2 V 2/3 ⎥⎦ . V1 V 5/3
V 5/3
V1
⎡⎛ V ⎞ 2/3 ⎤
3
1 ⎤ 3
2/3 ⎡ 1
Evaluating the limits give W = P1V1V1 ⎢ 2/3 − 2/3 ⎥ = P1V1 ⎢⎜ 1 ⎟ − 1⎥ . Using 2
V1 ⎦ 2
⎣ V2
⎢⎣⎝ V2 ⎠
⎥⎦
P1V1 = nRT1 and 3
J ⎞
V1 10L
2/3
⎛
=
= 2 , W = (1mol ) ⎜ 8.314
⎟⎠ ( 300K ) ⎡⎣( 2 ) − 1⎤⎦ = 2197J ⎝
2
K i mol
V2 5L
Verification: Use V1T1
f /2
= V2T2
f /2
⎛V ⎞
→ T2 = ⎜ 1 ⎟
⎝ V2 ⎠
2/ f
T1 with f = 3 for helium. This gives 2/3
f
f
3
⎛ 10L ⎞
T2 = ⎜
300K = 476K . Now use U = nRT → ΔU = nRΔT = nR (T2 − T1 ) ⎟
⎝ 5L ⎠
2
2
2
3
ΔU == (1mol ) 8.314J i K −1 i mol ( 476K − 300K ) = 2197J 2
Hence ΔU = W QED Final Note It is useful to compare the work done by compressing the gas from V1 = 10L to
V2 = 5L . As mentioned the magnitude of the work done is simply the area under the P vs. V curve for the interval from V1 = 10L to V2 = 5L . Isotherm Adiabat Wisotherm = 1730J Wadiabat = 2197J P3 P2 (
)
P1 P1 V1 V2 V1 V2 Discusssion: It is clear that Wadiabat > Wisotherm since the area under the P vs. V plot is greater. A more physical explanation is that for the isotherm, as work is being done to compress the gas heat escapes (Q < 0), decreasing the pressure of the gas, and consequently the amount of work required. On the other hand, in the adiabatic process there is no heat loss, and as the gas is compressed the pressure of the gas increases, making it harder to compress, which increases the amount of work needed to compress the gas further.