General Physics III Practice Exam I Solutions Fall 2007

3. Two samples of an ideal gas are initially at the same temperature and pressure. They are each compressed
reversibly from volume V to volume V/2, one isothermally, the other adiabatically.
(a) In which sample is the final pressure greater?
(b) Determine the change in entropy of the gas for each process.
(c) What is the entropy change of the environment for each process?
(22 points)
Solution:
(a) For the isothermal process we find the ratio of pressures from the ideal gas equation:
(P 2 V 2 /P 1 V 1 ) = T 2 /T 1 = 1, so P 2 /P 1 = V 1 /V 2 = 2.
(4 points)
For the adiabatic process,
P 1 V γ 1 = P 2 V γ 2 , so P 2 /P 1 = (V 1 /V 2 ) γ = 2 γ .
(4 points)
Because γ > 1, P2/P1 > 2 in the adiabatic case, so the final pressure is greater for the adiabatic process.
(b)
For the isothermal process we have
ΔE int = 0; Q = -W.
Thus ΔS = -W/T = [nRT ln(V 2 /V 1 )]/T = nR ln(1/2) = – nR ln 2.
(4 points)
For the adiabatic process we have
Q = 0.
Thus ΔS a = 0. (3 points)
(c)
Because each process is reversible, the energy change of the universe is zero.
For the isothermal process we have
ΔS surr = – ΔSi = nR ln 2. (4 points)
For the adiabatic process we have
ΔS surr = –ΔS = 0. (3 points)