Paper: CSIR-NET December 2019 · Subject: Physical Chemistry · Chapter: Gaseous State · Topic: Gaseous State – General · Marks: 2 · Difficulty: Medium
A monoatomic perfect gas undergoes expansion from $(p_1, V_1)$ to $(p_2, V_2)$ under isothermal or adiabatic conditions. The pressure of the gas will fall more rapidly under adiabatic conditions because
(a)$p \propto \frac{1}{V}$
(b)$p \propto \frac{1}{V^{7 / 5}}$
(c)$p \propto \frac{1}{V^{3 / 2}}$
(d)$p \propto \frac{1}{V^{5/3}}$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
For a reversible adiabat pV^γ = constant, with γ = 5/3 for a monatomic gas. p ∝ V^(−5/3) falls faster than the isothermal p ∝ 1/V (d).