A mole of gas undergoes an expansion through two different processes. The gas satisfies the equation of state pV = NR T where R is a constant, N the number of moles, p the pressure, T the temperature and V the volume of the gas. The initial and final temperatures are T_0. The walls of the gas container are diathermal. However, if a process takes place extremely fast, the walls can be considered as adiabatic. The initial pressure of the gas is p_1, the final pressure is p_2. Express the work performed on the gas in terms of p_1, p_2 and T_0 for the following processes:
a) a reversible isothermal process,
b) an extremely fast pressure variation, such that the external pressure on the gas is p_2 during the expansion, then an isochoric process during which the temperature again reaches the equilibrium temperature T_0,
Question 2.7: A mole of gas undergoes an expansion through two different p...
To compute the work performed on the gas undergoing a process from the initial state (V_1, p_1, T_0) to the final state (V_2, p_2, T_0), we use the state equation,
p_1 V_1 = p_2 V_2 = N R T_0 thus \frac{V_2}{V_1} =\frac{P_1}{P_2}.
a) The work performed on the gas during a reversible expansion from the initial state (V_1, p_1, T_0) to the final state (V_2, p_2, T_0) is obtained by calculating the integral expression (2.42),
W_{if} = \int_{i}^{f}{\delta W} = -\int_{V_i}^{V_f}{p (S,V)dV} (reversible process) (2.42)
W_{12} = – \int_{V_1}^{V_2}{p dV} = -N R T_0 \int_{V_1}^{V_2}{\frac{dV}{V} } = -N R T_0 \ln \Bigl(\frac{V_2}{V_1}\Bigr) = -N R T_0 \ln \Bigl(\frac{p_1}{p_2}\Bigr) =N R T_0 \ln \Bigl(\frac{p_2}{p_1}\Bigr) .
b) There is no work performed on the gas during an isochoric process since the volume is constant. However, to determine the work performed by the environment at pressure p^{ ext} = p_2 on the gas during an irreversible isobaric process from the initial state (V_1, p_1, T_0) to the final state (V_2, p_2, T_0), we need to integrate the general expression (2.36) for the mechanical power P_W over time,
P_W = −p^{ ext}\dot{V} (2.36)
W_{12} = \int_{t_1}^{t_2}{P_W dt} = – \int_{V_1}^{V_2}{p^{ext}dV } = – \int_{V_1}^{V_2}{p_2}dV = -p_2 \int_{V_1}^{V_2}{dV} =-p_2 V_2 +p_2 V_1 = \frac{p_2}{p_1} (p_1V_1)-p_2 V_2 = N R T_0\Bigl(\frac{p_2}{p_1}-1\Bigr).