Reversible isobaric transformation

The reversible isobaric transformation is the transformation at constant pressure: $P_B = P_A$. The piston is free to move but the load resting on it does not change, so the pressure of the gas stays fixed at its value while volume and temperature vary together.

To derive the entropy change it is convenient to exploit the ideal gas law. At constant pressure the ratio between volume and temperature is fixed, $V_B/V_A = T_B/T_A$, and so the two logarithms coincide:

$\ln\frac{V_B}{V_A} = \ln\frac{T_B}{T_A}$

Substituting into the general formula, the two contributions — the one in $C_v$ and the one in $R$ — add up within the same logarithm. Recalling that $C_p = C_v + R$ for an ideal gas:

$\Delta S_\text{gas} = (nC_v + nR)\,\ln\frac{T_B}{T_A} = nC_p\,\ln\frac{T_B}{T_A}$

The work is that characteristic of a constant-pressure transformation: the force on the piston is constant and the gas multiplies it by the volume displacement, $L_\text{gas} = P(V_B - V_A)$. Using the ideal gas law again, $P\,\Delta V = nR\,\Delta T$. The heat absorbed, at constant pressure, is governed by the heat capacity $C_p$: $Q = nC_p(T_B - T_A)$.

Principle — Reversible isobaric transformation

$\begin{aligned} P_B &= P_A \\ L_\text{gas} &= P(V_B - V_A) = nR(T_B - T_A) \\ Q &= nC_p(T_B - T_A) \\ \Delta S_\text{gas} &= nC_p\,\ln\frac{T_B}{T_A} \end{aligned}$

The energy flow is nicely visualised with a bubble diagram: heat $Q$ enters the internal energy of the gas from the ambient reservoir, which in turn hands work $L_\text{gas}$ to the piston, which finally pushes away the atmosphere by doing work $L_\text{atm}$.

Bubble diagram of a reversible isobaric transformation: heat $Q$ enters the internal energy of the gas from the reservoir, which in turn does work $L_\text{gas}$ on the piston, which in turn pushes away the atmosphere ($L_\text{atm}$).

Warning — $L_\text{gas} = P\,\Delta V$ only holds if reversible

The formula $L_\text{gas} = P_\text{gas}\,\Delta V_\text{gas}$ requires the pressure of the gas to be the external pressure on the piston at every instant. This is true only in a reversible transformation. In a free expansion, for example, the gas exerts no pressure against anything ($P_\text{esterna} = 0$) so $L_\text{gas} = 0$ even if $\Delta V > 0$.

Connections

Topics: Thermodynamics Concepts: Thermodynamic transformations · Entropy · Ideal gas law · First law of thermodynamics Skills: Solving a thermodynamic cycle Objects: Ideal gas · Piston and cylinder

Related exercises: Worked exercise — reversible vs irreversible isothermal expansion · Problem — Rectangular cycle in the p-V plane · Isothermal expansion: ΔS of the gas