# What is the change in entropy for an adiabatic expansion against an external pressure for an ideal gas?

May 15, 2017

I assume you mean the entropy of the system. This won't be zero since the volume of the gas changes, assuming the expansion is reversible and that the gas is ideal.

If we are to examine $\Delta {S}_{s y s}$ for an adiabatic expansion, what we should recall is that:

• Reversible work is ${w}_{r e v} = - {\int}_{{V}_{1}}^{{V}_{2}} P \mathrm{dV}$
• Reversible heat flow is $\delta {q}_{r e v} = T \mathrm{dS}$

In an expansion, the change in volume is positive, so that ${w}_{r e v} < 0$. If it's adiabatic, the heat flow between the system and surroundings is zero, i.e. $\delta {q}_{r e v} = 0$.

By the first law of thermodynamics:

$\Delta U = {\cancel{{q}_{r e v}}}^{0} + {w}_{r e v} = - P \int \mathrm{dV}$,

or in differential form:

$\mathrm{dU} = {\cancel{\delta {q}_{r e v}}}^{0} + \delta {w}_{r e v} = - P \mathrm{dV}$, $\text{ "" } \boldsymbol{\left(1\right)}$

where $U$ is the internal energy.

So all the work goes into changing the internal energy of the system, while $\Delta S$ due to reversible heat flow is zero.

However, the change in entropy can also be written as a function of temperature $T$ and volume $V$:

$\mathrm{dS} \left(T , V\right) = {\left(\frac{\partial S}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$$\text{ "" } \boldsymbol{\left(2\right)}$

This treats the change in entropy as two steps:

1. Change in entropy due to a change in temperature at a constant volume.
2. Change in entropy due to a change in volume at a constant temperature.

Note that the second term involves a change in entropy due to a change in volume, which is related to the work. With some manipulation using Maxwell's Relations, we end up with:

$\mathrm{dS} \left(T , V\right) = \frac{1}{T} {\left(\frac{\partial U}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$ $\text{ "" } \boldsymbol{\left(3\right)}$

where $P$ is pressure, and $T$ is temperature.

For an ideal gas, we can evaluate using the ideal gas law that

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{\partial}{\partial T} {\left[\frac{n R T}{V}\right]}_{V} = \frac{n R}{V}$,

so:

$\mathrm{dS} \left(T , V\right) = \frac{1}{T} {\left(\frac{\partial U}{\partial T}\right)}_{V} \mathrm{dT} + \frac{n R}{V} \mathrm{dV}$ $\text{ "" } \boldsymbol{\left(4\right)}$

Note that in an adiabatic process, we have that $\mathrm{dU} = - P \mathrm{dV}$ from $\left(1\right)$, so dividing by $\mathrm{dT}$ at a constant volume:

${\left(\frac{\partial U}{\partial T}\right)}_{V} = - P {\cancel{{\left(\frac{\partial V}{\partial T}\right)}_{V}}}^{0} = 0$

Treating the change in entropy in two steps, integrating $\left(4\right)$ over a temperature range for the first step and volume range for the second step:

$\Delta S \left(T , V\right) = {\int}_{{T}_{1}}^{{T}_{2}} \frac{1}{T} {\cancel{{\left(\frac{\partial U}{\partial T}\right)}_{V}}}^{0} \mathrm{dT} + {\int}_{{V}_{1}}^{{V}_{2}} \frac{n R}{V} \mathrm{dV}$

But the integral of zero is zero (no area under the curve!), so:

$\textcolor{b l u e}{\Delta S \left(T , V\right)} = 0 + \textcolor{b l u e}{n R \ln \left({V}_{2} / {V}_{1}\right)}$

Thus, $\Delta S \ne 0$ for an adiabatic expansion against an external pressure for an ideal gas.