# Physical Chemistry/Thermodynamic Processes for an Ideal Gas
## Thermodynamic Processes for an Ideal Gas

Let us first consider the expansion and compression of an ideal gas from an initial volume V$_1$ to a final volume V$_2$ under constant-temperature (isothermal) conditions. From the definition of pressure-volume work, we have: $w=-\int _{V_{1}}^{V_{2}}P_{ext}dV$ If we assume the process occurs reversibly, then $P_{ext}=P$ where P is the pressure of the gas; therefore $w=-\int _{V_{1}}^{V_{2}}PdV$ Now let us use the equation of state for an ideal gas, which is $PV=nRT$ where n is the number of moles of gas, R is the ideal gas constant, and T is the absolute (Kelvin) temperature. Substituting for P, we get $w=-\int _{V_{1}}^{V_{2}}{\frac {nRT}{V}}dV$ If we have a closed system, then n is a constant; moreover, since we have stipulated that this process occurs under isothermal conditions, T is also a constant. Therefore, the integral simplifies to: $w=-nRT\int _{V_{1}}^{V_{2}}{\frac {dV}{V}}$ This integral may be solved to yield $w=-nRT\ln \left({\frac {V_{2}}{V_{1}}}\right)\;$
Next we consider the change in the internal energy of the ideal gas upon isothermal expansion/compression. It is helpful to review the two key assumptions that underlie the ideal gas model:
* Ideal gas particles are point particles.
* Ideal gas particles have no interparticular forces.
From these two statements, we now contemplate what would happen to the internal energy (U) of an ideal gas upon isothermal expansion or compression. Because ideal gas particles are point particles, if the gas sample is compressed, the particles will not experience any change in energy because they have no internal volume to get "squeezed". Similarly, because they have no interparticular forces, they do not experience any change in attraction or repulsion between different particles if they are squeezed closer together. Hence, we may (correctly) reason that for an isothermal process on an ideal gas, the change in internal energy is zero. $\Delta U=0\;$ (isothermal process, ideal gas) (This same result may be rigorously proved using the thermodynamic master equations, or by using statistical thermodynamics.) Note that because U is a state function, $\Delta U$ will always be equal to zero for any isothermal process on an ideal gas, whether or not it occurs reversibly. Because of the First Law of Thermodynamics, $\Delta U=q+w\;$ and because $\Delta U=0$, the energy change associated with heat upon isothermal reversible expansion or compression is just the negative of the work, or $q=-w=nRT\ln {\frac {V_{2}}{V_{1}}}$ . In a similar fashion, other results may be derived for processes on ideal gases. A summary of results is given below.

### Equation Table ($PV^{m}=constant$)

| | Constant Pressure | Constant Volume | Isothermal | Adiabatic |
| Variable | $\Delta P=0\;$ | $\Delta V=0\;$ | $\Delta T=0\;$ | $q=0\;$ |
| $m\;$ | $0\;$ | $\infty \;$ | $1\;$ | $\gamma ={\frac {C_{p}}{C_{v}}}={\frac {5}{3}}\;$ [for a monoatomic ideal gas] |
| Work ${\begin{matrix}w=-\int _{V_{1}}^{V_{2}}PdV\end{matrix}}$ | $-P\left(V_{2}-V_{1}\right)\;$ | $0\;$ | $-nRT\ln \left({\frac {V_{2}}{V_{1}}}\right)\;$ | $C_{v}\left(T_{2}-T_{1}\right)\;$ |
| Heat Capacity, $C\;$ | $C_{p}=(5/2)nR\;$ | $C_{v}=(3/2)nR\;$ | $\infty \;$ | $C_{p}\;$ or $C_{v}\;$ |
| Internal Energy, $\Delta U=q+w\;$ | $q+w\;$ $q_{p}+P\Delta V\;$ | $q\;$ $w=0\;$ $C_{v}\left(T_{2}-T_{1}\right)\;$ | $0\;$ $q=-w\;$ | $w\;$ $q=0\;$ $C_{v}\left(T_{2}-T_{1}\right)\;$ |
| Enthalpy, $\Delta H\;$ $H=U+PV\;$ | $C_{p}\left(T_{2}-T_{1}\right)\;$ | $q_{v}+V\Delta P\;$ | $0\;$ | $C_{p}\left(T_{2}-T_{1}\right)\;$ |
| Entropy ${\begin{matrix}\Delta S=-\int _{T_{1}}^{T_{2}}{\frac {C}{T}}dT\end{matrix}}$ | $C_{p}\ln \left({\frac {T_{2}}{T_{1}}}\right)\;$ | $C_{v}\ln \left({\frac {T_{2}}{T_{1}}}\right)\;$ | $nR\ln \left({\frac {V_{2}}{V_{1}}}\right)\;$ | $0\;$ |

## Molar Thermodynamic Properties for an Ideal Gas

We are given that the molar Gibbs free energy of an ideal is ${\bar {G}}={\bar {G}}^{\circ }+RT\ln {\frac {P}{P^{\circ }}}\;$. Let's try and derive ${\bar {U}}={\bar {U}}^{\circ }={\bar {H}}^{\circ }-RT\;$; noting that ${\bar {U}}^{\circ }={\bar {G}}^{\circ }+T{\bar {S}}^{\circ }-RT\;$ and assuming that the internal energy $U\;$ and enthalpy $H\;$ of an ideal gas are independent of pressure and volume.