Simulation · Physics · Class 11

Isothermal or adiabatic: same gas, two paths

From the lesson Adiabatic process in Thermodynamics. Change the values and watch what happens.

The idea behind it

NCERT §11.8.3

In an adiabatic process the system is insulated, so no heat enters or leaves: ΔQ = 0. The first law then gives ΔW = −ΔU: work done by the gas comes out of its internal energy, so an ideal gas cools as it expands adiabatically.
For an ideal gas undergoing a quasi-static adiabatic change, PV^γ = constant, where γ = C_p/C_v (ratio of the specific heats, ordinary or molar). The result is quoted without proof.
So between two states: P₁V₁^γ = P₂V₂^γ. Since γ > 1, an adiabat is steeper than an isotherm through the same point: for the same compression the pressure rises more.
Fig. 11.8: two adiabats connect two isotherms on a P–V diagram.
Work done by the gas from (P₁, V₁, T₁) to (P₂, V₂, T₂): W = ∫P dV with P = constant/V^γ, giving W = [P₁V₁ − P₂V₂]/(γ − 1) = μR(T₁ − T₂)/(γ − 1).
If the gas does work (W > 0), T₂ < T₁: it cools. If work is done on it (W < 0), T₂ > T₁: it warms up.
Example: a gas with γ = 1.4 (such as hydrogen) compressed adiabatically to half its volume: P₂/P₁ = 2^1.4 ≈ 2.64, while an isothermal halving only doubles the pressure.
For the same case, TV^(γ−1) = constant gives T₂ = T₁ × 2^0.4 ≈ 1.32 T₁: 300 K becomes about 396 K.

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