Reversible isochoric transformation

The reversible isochoric transformation is obtained by imposing the constraint of constant volume: $V_B = V_A$. In practice the piston is locked, or the gas is enclosed in a rigid container. Since the volume does not change, the gas does not move any boundary and therefore does no work on the piston: $L_\text{gas} = 0$.

For the transformation to be reversible, however, the gas cannot be heated suddenly by contact with a single, much hotter reservoir: it must exchange heat with a continuous sequence of reservoirs, each at a slightly different temperature from the next, so as to remain in thermal equilibrium at every instant. All the absorbed heat then goes into increasing the internal energy of the gas, since none of it can be converted into work.

Applying the general entropy formula, the volume-related term vanishes because $\ln(V_B/V_A) = \ln 1 = 0$, and only the thermal contribution remains:

$\Delta S_\text{gas} = nC_v\,\ln\frac{T_B}{T_A}$

The heat exchanged is what, at constant volume, changes the internal energy, i.e. $Q = nC_v(T_B - T_A)$.

Principle — Reversible isochoric transformation

$\begin{aligned} V_B &= V_A \\ L_\text{gas} &= 0 \\ Q &= nC_v(T_B - T_A) \\ \Delta S_\text{gas} &= nC_v\,\ln\frac{T_B}{T_A} \end{aligned}$

Connections

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

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