1. The entropy production rate is given by (2.37):
Π_S=\frac{1}{T(S,V)}\Bigl(p(S,V)-p^{ext}\Bigl)\dot{V} \geq 0.
Π_S=\frac{\dot{V}}{T}(p-p^{ext})\geq 0.
Initially, before the additional mass \acute{M}is added, the system consisting of the piston and the gas is at equilibrium. This implies that the initial pressure p_i of the ideal gas (5.47)
pV = NR T
is equal to the pressure p^{ext}_i exerted by the piston (since the atmospheric pressure is negligible),
p_i = p^{ext}_i where p_i =\frac{NR T_i}{V_i} and p^{ext}_i =\frac{Mg}{A}.
This implies that,
NR T_i=\frac{Mg}{A}V_i.
Finally, after the compression, when the piston is at rest, the system consisting of the piston and the gas is at equilibrium. This means that the final pressure p_f of the ideal gas is then equal to the final pressure p^{ext}_ f exerted by the piston and the additional mass,
p_f=p^{ext}_ f where p_f =\frac{NR T_f}{V_f} and p^{ext}_f =\frac{(M+\acute{M})g}{A}.
This implies,
NR T_f=\frac{(M+\acute{M})g}{A}V_f.
Thus, before and after the adiabatic compression, the system is at equilibrium and the entropy production rate vanishes, i.e. ΠS = 0. However, during the adiabatic compression, when p_i < p < p_f and p^{ext} = p^{ext}_ f , the entropy production rate is positive, i.e. ΠS > 0 as p < p^{ext} for \dot{V} < 0 and T > 0.
This implies that the adiabatic compression is irreversible.
2. As the compression is adiabatic, the heat provided to the system vanishes, i.e. Q_{if} = 0.
Thus, according to the first law (1.44), the internal energy variation of the gas is only due to the work W_{if} performed on the gas by the weight of the piston and the additional mass,
ΔU_{if} = W_{if }+ Q_{if}. (closed system)
ΔU_if = W_if.
The internal energy variation is given by,
ΔU_if = cNR(T_f − T_i).
Taking into account the fact that the pressure is constant during the compression process, the work W_if performed on the gas by the weight of the piston and of the additional mass is expressed as,
W_if = −p^{ ext}_f (V_f − V_i).
Thus, according to the first law,
cNR(T_f − T_i)=-\frac{(M+\acute{M})g}{A}(V_f-V_i).
This can be expressed in terms of the initial and final equilibrium states, respectively,
c\Bigl((M+\acute{M})V_f-MV_i\Bigr)=(M+\acute{M})(V_f-V_i).
Finally, as V_i = h_i A and V_f = h_f A, the piston height ratio yields,
\frac{h_f}{h_i} =\frac{V_f}{V_i}=\frac{1+\frac{1}{c+1}\frac{\acute{M} }{M} }{1+\frac{\acute{M}}{M} } \lt 1.
3. According to expression (5.88) for the ideal gas, the entropy variation ΔS_{if} reads,
ΔS_{if} = (c + 1) NR\ln \Bigl(\frac{V_f}{V_i} \Bigr)+cNR\ln \Bigl(\frac{p_f}{p_i} \Bigr).
Taking into account that,
\frac{p_f}{p_i}=\frac{p^{ext}_f}{p^{ext}_i}=1+\frac{\acute{M} }{M}.
the entropy variation ΔS_if for this irreversible adiabatic process is,
ΔS_{if} = (c + 1) NR\ln \Bigl(\frac{1+\frac{1}{c+1}\frac{\acute{M} }{M} }{1+\frac{\acute{M} }{M}} \Bigr)+cNR\ln \Bigl(1+\frac{\acute{M} }{M}\Bigr)=NR\Biggl(\frac{\bigl(1+\frac{1}{c+1}\frac{\acute{M} }{M} \bigr)^{c+1}}{1+\frac{\acute{M} }{M}} \Biggr)\gt 0.
The last step is obtained by performing a series expansion of the numerator and denominator.
4. When the pressure p of the gas is equal to the outside pressure p^{ext}, the entropy production rate vanishes, i.e.
ΠS = 0
Thus, the process is reversible. As the process is also adiabatic, i.e. P_Q = 0, the relation (2.40) implies that it is isentropic,
P_Q = T(S, V)\dot{S} (reversible process)
\dot{S}=0.
5. For an isentropic process (5.90), pressure and volume satisfy the following condition,
pV^γ = const.
p_i V^γ_i== p_f V^γ_f.
Thus, the volume ratio is,
\frac{V_f}{V_i}=\biggl(\frac{p_i}{p_f}\biggr)^{1/γ}.
For a reversible compression, the outside pressures p^{ext}_i and p^{ext}_f are equal to the pressures p_i and p_f of the gas, respectively. Thus,
\frac{p_i}{p_f}=\frac{p^{ext}_i}{p^{ext}_f}=\frac{1}{1+\frac{\acute{M} }{M}}.
Therefore, the height ratio, which is equal to the volume ratio, can be written as,
\frac{h_f}{h_i}=\frac{V_f}{V_i}=\Bigl(\frac{p_i}{p_f} \Bigr)^{1/\gamma }=\biggl(\frac{1}{1+\frac{\acute{M} }{M} } \biggr)^{1/\gamma }<1.
6. According to equation (5.88) for the ideal gas, the entropy variation ΔS_{if} for a reversible adiabatic process vanishes,
ΔS_{if} = (c + 1) NR\ln \Bigl(\frac{V_f}{V_i} \Bigr)+cNR\ln \Bigl(\frac{p_f}{p_i} \Bigr).
ΔS_{if }= (c + 1) NR\ln \Biggl(\biggl(\frac{p_i}{p_f} \biggr)^{1/\gamma }\Biggr)+cNR\ln \biggl(\frac{p_f}{p_i} \biggr)=0.
This exercise illustrates the point that the irreversibility of a process can be related to the difference in pressure between a system and its environment. Likewise, in § 3.2, we found that irreversibility was related to the difference in temperature between two subsystems, one of which could be considered as the environment of the other.
