Gibbs Energy of an Ideal Gas — Theory
For an ideal gas undergoing an isothermal change in pressure from the
standard reference pressure \(P^{\circ}\) (usually 1 bar) to a pressure \(P\),
the change in Gibbs energy depends only on the pressure ratio \(P/P^{\circ}\).
The natural logarithm is used throughout.
\[
\begin{aligned}
\Delta G & = nRT\,\ln\!\left(\frac{P}{P^{\circ}}\right) \\[4pt]
G &= G^{\circ} \;+\; nRT\,\ln\!\left(\frac{P}{P^{\circ}}\right) \\[4pt]
G_m &\equiv \frac{G}{n} \;=\; G_m^{\circ} \;+\; RT\,\ln\!\left(\frac{P}{P^{\circ}}\right)
\end{aligned}
\]
Where the equation comes from (sketch)
For an ideal gas in an isothermal process, \(\Delta H=0\).
Using \(\Delta G=\Delta H - T\Delta S\) and the entropy change for an
isothermal ideal-gas compression/expansion,
\(\Delta S = -\,nR\,\ln\!\big(P/P^{\circ}\big)\), we obtain
\[
\begin{aligned}
\Delta G
&= -\,T\Delta S
\;=\; -\,T\!\left[ -\,nR\,\ln\!\left(\frac{P}{P^{\circ}}\right)\right]
\;=\; nRT\,\ln\!\left(\frac{P}{P^{\circ}}\right).
\end{aligned}
\]
Physical meaning & sign
-
If \(P < P^{\circ}\), then \(\ln(P/P^{\circ}) < 0\) and
\(\Delta G < 0\): expansion to the lower pressure lowers \(G\).
-
If \(P > P^{\circ}\), then \(\ln(P/P^{\circ}) > 0\) and
\(\Delta G > 0\): compression raises \(G\).
-
The molar relation can be written as the chemical potential:
\(\mu \equiv G_m = \mu^{\circ} + RT\,\ln(P/P^{\circ})\).
What the calculator does
-
Converts the entered pressures to a common base (Pa) so the ratio
\(P/P^{\circ}\) is dimensionless. Any consistent pressure units are accepted.
-
Converts temperatures to kelvin and, when needed, \(G^{\circ}\) or \(G_m^{\circ}\)
to joules (or joules per mole).
-
Evaluates the chosen relation with \(R=8.314\,462\,618\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
Assumptions & limits
- Ideal-gas behavior; deviations at high \(P\) or very low \(T\) are not included.
- Isothermal process (constant \(T\)).
-
\(P^{\circ}\) is a reference state (1 bar by convention).
Only the ratio \(P/P^{\circ}\) matters; absolute units cancel.
Sanity check example
With \(n=2.0\ \mathrm{mol}\), \(T=298.15\ \mathrm{K}\), \(P=5.0\ \mathrm{bar}\),
\(P^{\circ}=1.0\ \mathrm{bar}\),
\[
\Delta G = nRT\,\ln(5) \approx 2.0\times 8.314\times 298.15\times 1.609
\approx 7.98\times 10^{3}\ \mathrm{J} = 7.98\ \mathrm{kJ}.
\]
