General Gas Equation — Theory & Guide
For the same amount of gas (constant \(n\)) changing from an initial state (i) to a final state (f),
applying \(PV=nRT\) to both states and dividing gives the
general gas equation (combined gas law):
\[
\frac{P_i V_i}{T_i}=\frac{P_f V_f}{T_f}
\]
Derivation (brief)
Write \(PV=nRT\) for each state with the same \(n\) and \(R\); then divide:
\[
\begin{aligned}
P_iV_i &= nRT_i \\
P_fV_f &= nRT_f \\
\frac{P_iV_i}{nRT_i} &= \frac{P_fV_f}{nRT_f} \\
\Rightarrow\quad \frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f}.
\end{aligned}
\]
Symbols
- \(P\): absolute pressure (Pa, kPa, bar, atm, mmHg/torr).
- \(V\): volume (m³, L, mL, cm³).
- \(T\): temperature in Kelvin. Convert with \(T_\text{K}=t_{^\circ\!C}+273.15\) or \(T_\text{K}=(t_{^\circ\!F}-32)\tfrac{5}{9}+273.15\).
Use absolute pressure and \(T>0\ \mathrm{K}\).
Rearrangements (solve for any one)
Final pressure \(P_f\)
\[
\begin{aligned}
\frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f} \\
P_f &= \frac{P_iV_i}{T_i}\,\frac{T_f}{V_f}
\end{aligned}
\]
Final volume \(V_f\)
\[
\begin{aligned}
\frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f} \\
V_f &= \frac{P_iV_i}{T_i}\,\frac{T_f}{P_f}
\end{aligned}
\]
Final temperature \(T_f\)
\[
\begin{aligned}
\frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f} \\
T_f &= \frac{P_fV_f}{P_iV_i}\,T_i
\end{aligned}
\]
Initial pressure \(P_i\)
\[
\begin{aligned}
\frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f} \\
P_i &= \frac{P_fV_f}{T_f}\,\frac{T_i}{V_i}
\end{aligned}
\]
Initial volume \(V_i\)
\[
\begin{aligned}
\frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f} \\
V_i &= \frac{P_fV_f}{T_f}\,\frac{T_i}{P_i}
\end{aligned}
\]
Initial temperature \(T_i\)
\[
\begin{aligned}
\frac{P_iV_i}{T_i} &= \frac{P_fV_f}{T_f} \\
T_i &= \frac{P_iV_i}{P_fV_f}\,T_f
\end{aligned}
\]
Special cases
- Boyle’s law (isothermal \(T\) constant): \(P_1V_1=P_2V_2 \Rightarrow P\propto \tfrac1V\).
- Charles’s law (isobaric \(P\) constant): \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \Rightarrow V\propto T\).
- Amontons/Gay-Lussac (isochoric \(V\) constant): \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2} \Rightarrow P\propto T\).
Worked example
A 2.50 L sample at \(98.4\ \text{kPa}\) and \(24^\circ\text{C}\) is cooled to \(-25^\circ\text{C}\) at the same pressure.
Find \(V_f\).
Plan. Pressure is constant, so \(\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}\Rightarrow V_f=V_i\,\dfrac{T_f}{T_i}\).
\[
\begin{aligned}
T_i &= 24 + 273.15 = 297.15\ \text{K} \\
T_f &= -25 + 273.15 = 248.15\ \text{K} \\
V_f &= 2.50\ \text{L}\times \frac{248.15}{297.15} = 2.09\ \text{L}\ \text{(3 s.f.)}
\end{aligned}
\]