AcademyFirst-Law Thermodynamics
Academy
Internal Energy of Ideal Gases
Level 1 - Physics topic page in First-Law Thermodynamics.
Principle
For an ideal gas, internal energy depends only on temperature.
Notation
\(U\)
internal energy
\(\mathrm{J}\)
\(n\)
amount of gas
\(\mathrm{mol}\)
\(C_V\)
molar heat capacity at constant volume
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(T\)
absolute temperature
\(\mathrm{K}\)
\(R\)
molar gas constant
\(\mathrm{J\,mol^{-1}\,K^{-1}}\)
\(f\)
active quadratic degrees of freedom
1
Method
Derivation 1: Link energy to temperature
In the ideal-gas model, microscopic kinetic energy sets temperature. Changing volume at fixed temperature does not change internal energy.
Temperature dependence
\[U=U(T)\]
Heat-capacity definition
\[dU=nC_V\,dT\]
Finite change
\[\Delta U=nC_V\Delta T\]
Derivation 2: Use the monatomic model
For a monatomic ideal gas, the three translational degrees of freedom give the internal energy.
Equipartition form
\[U=\frac{f}{2}nRT\]
Monatomic gas
\[f=3\]
Monatomic energy
\[U=\frac{3}{2}nRT\]
Rules
These are the ideal-gas internal-energy relations.
Ideal-gas change
\[\Delta U=nC_V\Delta T\]
Equipartition model
\[U=\frac{f}{2}nRT\]
Monatomic gas
\[U=\frac{3}{2}nRT\]
Monatomic change
\[\Delta U=\frac{3}{2}nR\Delta T\]
Examples
Question
Find
\[\Delta U\]
for \[2.0\,\mathrm{mol}\]
of monatomic ideal gas warmed by \[40\,\mathrm{K}\]
Answer
\[\Delta U=\frac{3}{2}nR\Delta T=\frac{3}{2}(2.0)(8.31)(40)=1.0\times10^3\,\mathrm{J}\]
Checks
- Use Kelvin temperature changes.
- For an ideal gas, \(\\Delta U\) depends on \(\\Delta T\), not directly on \(\\Delta V\).
- Monatomic ideal gas means \(C_V=3R/2\).
- If \(T\) is unchanged, \(\\Delta U=0\) for an ideal gas.