AcademyFirst-Law Thermodynamics
Academy
Heat Capacity of Ideal Gases
Level 1 - Physics topic page in First-Law Thermodynamics.
Principle
An ideal gas needs more heat at constant pressure than at constant volume.
Notation
\(C_V\)
molar heat capacity at constant volume
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(C_P\)
molar heat capacity at constant pressure
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(\gamma\)
heat-capacity ratio
1
\(n\)
amount of gas
\(\mathrm{mol}\)
\(R\)
molar gas constant
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(\Delta T\)
temperature change
\(\mathrm{K}\)
Method
Derivation 1: Compare constant volume and constant pressure
At constant volume, no volume work is done. At constant pressure, heating also supplies expansion work.
Constant volume
\[Q_V=\Delta U=nC_V\Delta T\]
Constant pressure work
\[W=p\Delta V=nR\Delta T\]
First law at constant pressure
\[Q_P=\Delta U+W\]
Mayer relation
\[nC_P\Delta T=nC_V\Delta T+nR\Delta T\]
Derivation 2: Define the ratio
The ratio \(\\gamma\) appears in adiabatic ideal-gas processes and compares the two heat capacities.
Heat-capacity difference
\[C_P-C_V=R\]
Ratio
\[\gamma=\frac{C_P}{C_V}\]
Monatomic values
\[C_V=\frac32R,\qquad C_P=\frac52R,\qquad \gamma=\frac53\]
Rules
These are the ideal-gas heat-capacity relations.
Constant volume heat
\[Q_V=nC_V\Delta T\]
Constant pressure heat
\[Q_P=nC_P\Delta T\]
Mayer relation
\[C_P-C_V=R\]
Heat-capacity ratio
\[\gamma=\frac{C_P}{C_V}\]
Examples
Question
A gas has
\[C_V=20.8\,\mathrm{J\,mol^{-1}\,K^{-1}}\]
Find \(C_P\) and \(\gamma\).Answer
\[C_P=C_V+R=20.8+8.31=29.1\,\mathrm{J\,mol^{-1}\,K^{-1}}\]
\[\gamma=C_P/C_V=29.1/20.8=1.40\]
Checks
- \(C_P\) is larger because the gas can expand and do work.
- Molar heat capacities multiply by moles, not mass.
- \(C_P-C_V=R\) applies to ideal gases.
- \(\\gamma\) is dimensionless.