Internal Energy of an Ideal Gas

We will show that the internal energy of an ideal gas is a function of temperature only. This makes physical sense because there is an assumption in ideal gas behavior that there is no interaction between the molecules when we write $P \ensuremath{\overline{V}} = RT$

Start with a reversible process for an ideal gas:

$$\input{equations/72A}$$ (11-1)

Consider two processes: one occurring at constant volume, the other occurring at constant pressure.

Figure 11-1: Two consecutive processes, constant volume followed by constant pressure.

For process 1: $dU = C_V dT + 0$; This can be integrated because $T$ is the only thing that is changing on the righthandside ( $C_V$ is assumed to be independent of $T$ and $V$).

For process 2: $dU = C_P dT - PdV$; $P$ is constant (i.e., not a function of $T$ or $V$) so it can be integrated directly. Using the ideal gas law:

$$\input{equations/dpv}$$ (11-2)

So for process 2,

$$\input{equations/73A}$$ (11-3)

Since we can make up any quasi-static curve with segments of $dV$ processes and $dP$ processes

Figure 11-2: Any curve can be made up of short segments in the limit.

Evidently, the sum of any such processes is a function only of $T$. Therefore, for an ideal gas

$$\input{equations/73B}$$ (11-4)

Comparing the two equations:

$$\input{equations/74A}$$ (11-5)

for the constant volume process, and

$$\input{equations/74B}$$ (11-6)

for the constant pressure process:

$$\input{equations/74C}$$ (11-7)

$$\input{equations/74D}$$ (11-8)

$$\input{equations/74E}$$ (11-9)

$$\input{equations/CpCv}$$ (11-10)

or

$$\input{equations/CpCv-molar}$$ (11-11)