# Entropy and heat

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.28.

The thermodynamic identity for an infinitesimal process is

$\displaystyle dU=TdS-PdV \ \ \ \ \ (1)$

This relation bears a resemblance to heat plus work relation for the change in internal energy:

$\displaystyle dU=Q+W \ \ \ \ \ (2)$

In a quasistatic process, where a gas is being compressed slowly enough that the pressure has a chance to equalize throughout the volume of the gas at each stage, the work done in compressing the gas is ${-PdV}$ (this is positive, as the volume decreases in compression so ${dV<0}$). In that case, then

 $\displaystyle W$ $\displaystyle =$ $\displaystyle -PdV\ \ \ \ \ (3)$ $\displaystyle Q$ $\displaystyle =$ $\displaystyle TdS \ \ \ \ \ (4)$

and the change in entropy can be calculated as

$\displaystyle dS=\frac{Q}{T} \ \ \ \ \ (5)$

which agrees with the original definition of entropy.

As an example, suppose we have a litre of air at room temperature (300 K) and atmospheric pressure (${10^{5}\mbox{ N m}^{-2}}$), and we heat it at constant pressure until it doubles in volume. From the ideal gas law, if ${P}$ is constant and ${V}$ doubles, then ${T}$ must also double. The entropy change is therefore

 $\displaystyle \Delta S$ $\displaystyle =$ $\displaystyle C_{P}\int_{T_{i}}^{T_{f}}\frac{dT}{T}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle C_{P}\ln\frac{T_{f}}{T_{i}}=C_{P}\ln2 \ \ \ \ \ (7)$

where ${C_{P}}$ is the heat capacity at constant pressure. From the appendix to Schroeder’s book, ${C_{P}\approx29\mbox{ J K}^{-1}}$ for one mole of air (the values for nitrogen and oxygen are both around 29). The number of moles of air in one litre is

 $\displaystyle n$ $\displaystyle =$ $\displaystyle \frac{PV}{RT}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{10^{5}\times10^{-3}}{\left(8.314\right)\left(300\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.04\mbox{ mol} \ \ \ \ \ (10)$

The change in entropy is therefore

$\displaystyle \Delta S=\left(0.04\right)\left(29\right)\ln2=0.81\mbox{ J K}^{-1} \ \ \ \ \ (11)$