Isochoric and isobaric

Problem

In an isochoric process the volume $V$ is constant and the gas does no work: how does it then exchange energy? Compare with an isobaric process and derive Mayer’s relation.

Solution

Isochoric process. With $V$ constant the gas does no work: $L = p\,\Delta V = 0$. The first law gives

$Q = \Delta U + L = \Delta U = n\,c_V\,\Delta T$

so all the heat absorbed increases the internal energy (and the temperature). Energy enters only as heat, not as work.

Isobaric process. At constant pressure the gas expands, so $L = p\,\Delta V > 0$. The first law gives

$Q = \Delta U + L = n\,c_V\,\Delta T + p\,\Delta V$

Using $Q = n\,c_p\,\Delta T$ and, from the equation of state at constant $p$, $p\,\Delta V = nR\,\Delta T$:

$n\,c_p\,\Delta T = n\,c_V\,\Delta T + nR\,\Delta T$

Mayer’s relation. Dividing by $n\,\Delta T$ gives

$c_p - c_V = R$

The difference $c_p - c_V$ is precisely the expansion work required to heat the gas by the same $\Delta T$ at constant pressure: that is why the isobaric process requires more heat than the isochoric one.

$\ev{c_p - c_V = R}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes · Specific heat and heat capacity Objects: Ideal gas