AcademyFirst-Law Thermodynamics
Academy
Adiabatic Ideal-Gas Processes
Level 1 - Physics topic page in First-Law Thermodynamics.
Principle
An adiabatic ideal-gas process changes temperature through work without heat transfer.
Notation
\(Q\)
heat transferred to the system
\(\mathrm{J}\)
\(W\)
work done by the system
\(\mathrm{J}\)
\(\Delta U\)
change in internal energy
\(\mathrm{J}\)
\(\gamma\)
heat-capacity ratio \(C_P/C_V\)
1
\(p,V,T\)
pressure, volume, absolute temperature
varies
\(n\)
amount of gas
\(\mathrm{mol}\)
Method
Derivation 1: Apply the first law
Adiabatic means no heat crosses the boundary. Expansion work therefore comes from internal energy.
Adiabatic condition
\[Q=0\]
First law
\[\Delta U=Q-W\]
Adiabatic result
\[\Delta U=-W\]
Ideal-gas change
\[\Delta U=nC_V\Delta T\]
Derivation 2: State the reversible adiabatic constraints
For a quasistatic reversible adiabatic ideal-gas process, pressure, volume, and temperature obey power-law constraints.
Pressure-volume law
\[pV^\gamma=\text{constant}\]
Temperature-volume law
\[TV^{\gamma-1}=\text{constant}\]
Work from energy
\[W=-\Delta U=nC_V(T_i-T_f)\]
Rules
These are the reversible adiabatic ideal-gas relations.
No heat transfer
\[Q=0\]
Energy-work link
\[\Delta U=-W\]
Adiabatic curve
\[pV^\gamma=\mathrm{constant}\]
Temperature-volume
\[TV^{\gamma-1}=\mathrm{constant}\]
Adiabatic work
\[W=nC_V(T_i-T_f)\]
Examples
Question
A monatomic ideal gas expands adiabatically and its temperature falls by
\[50\,\mathrm{K}\]
For \[1.5\,\mathrm{mol}\]
find work done by the gas.Answer
For monatomic gas,
\[C_V=3R/2\]
Since \[W=-\Delta U=nC_V(T_i-T_f)\]
\[W=(1.5)(\tfrac32)(8.31)(50)=9.35\times10^2\,\mathrm{J}\]
Checks
- Adiabatic means \(Q=0\), not \(\\Delta T=0\).
- During adiabatic expansion, an ideal gas cools.
- During adiabatic compression, an ideal gas warms.
- The power laws require a reversible quasistatic ideal-gas process.