a. In the initial state 1:
V^{\prime }_{1}=\frac{nRT_1}{P_1}=\frac{5\times 0.08206\times 300}{50}=2.46 liters
If the adiabatic expansion from 50 to 10 atm is carried out reversibly, then the process path follows PV_\gamma = constant, and in the final state 2:
V^{\prime }_{2}=\left\lgroup\frac{P_1V^{\gamma }_{1} }{P_2} \right\rgroup ^{1/\gamma } =\left\lgroup\frac{50\times 2.46^{5/3} }{10} \right\rgroup ^{3/5}=6.47 liters
and
T_2 =\frac{P_2V_2}{nR}=\frac{10\times 6.47}{5\times 0.08206}=158 K
For the irreversible process, which takes the gas from the state 1 to the state 3, as q = 0:
\Delta U^\prime =-w=-4000=nc_v(T_3-T_1)=5\times 1.5\times 8.3144\times (T_3-300)
and hence, T_3 = 236 K, which is higher than T_2 .
b. As the irreversible expansion from state 1 to state 3 was conducted adiabatically, no thermal energy was transferred into the system, and hence, the difference between the entropy at state 3 and the entropy at state 1 is the entropy created, \Delta S^{\prime }_{irr} , as a result of the irreversible process. This difference in entropy can be calculated by considering any reversible path from state 1 to state 3. Consider the reversible path 1\rightarrow a\rightarrow 3 shown in Figure 3.10, which is a reversible decrease in temperature from
300 to 236 K at constant volume, followed by a reversible isothermal expansion from V^{\prime }_{a} and V^{\prime }_{3}
For a reversible constant-volume process,
\delta q_v=nc_vdT=TdS^\prime
or
dS^\prime =\frac{nc_vdT}{T}
the integration of which, from state 1 to state a , gives
S^{\prime }_{a} -S^{\prime }_{1} =nc_v\ln \frac{T_a}{T_1}=5\times 1.5\times 8.3144\times \ln \frac{234}{300} =-15.0 J/K
For the reversible isothermal expansion from state a to state 3, as Δ Uʹ = 0:
q=w=nRT\ln \frac{V^{\prime }_{3} }{V^{\prime }_{a} }
where
V^{\prime }_{3} =\frac{nRT_3}{P_3}=\frac{5\times 0.08206\times 236}{10}=9.68 liters
and thus,
S^{\prime }_{3} -S^{\prime }_{a} =\frac{q}{T}=nR\ln \frac{V^{\prime }_{3} }{V^{\prime }_{a} } =5\times 8.3144\ln \frac{9.68}{2.46} =57.8 J/K
The entropy created during the irreversible expansion is thus
S^{\prime }_{3} -S^{\prime }_{1}=-15.0+57.0=42.0 J/K
Alternatively, the state of the gas could be changed from 1 to 3 along the path 1 → 2 → 3. As the reversible adiabatic expansion from state 1 to state 2 is isentropic:
S^{\prime }_{3} -S^{\prime }_{1}=S^{\prime }_{3} -S^{\prime }_{2}
and, for the reversible isobaric expansion from state 2 to state 3:
\delta q_p=nc_pdT=TdS^\prime
or
dS^\prime =\frac{nc_pdT}{T}
the integration of which, from state 2 to state 3, gives
S^{\prime }_{3} -S^{\prime }_{2}=5\times 2.5\times 8.3144\times \ln \frac{236}{158} =42.0 J/K
which, again, is the entropy created by the irreversible adiabatic expansion of the gas from state 1 to state 3.
