24. An ideal gas at pressure P_0, volume V_0 and temperature T_0 is allowed to
isothermally expand to twice its initial volume. What is the final pressure?
A) P_0/2
B) 2 P_0
C) P_0
D) Undetermined

Isothermal Expansion of an Ideal Gas

An ideal gas at initial pressure de>P₀, volume de>V₀, and temperature de>T₀ undergoes isothermal expansion to final volume de>2V₀. Since temperature remains constant, Boyle’s law applies:

Boyle’s Law: de>P₁V₁ = P₂V₂

For the initial state, de>P₀V₀ = P_f(2V₀), giving the final pressure:

Final Pressure: de>P_f = P₀V₀ / (2V₀) = P₀ / 2

Step-by-Step Solution

Initial conditions: de>P₀, V₀, T₀
Final volume: de>V_f = 2V₀
Final temperature: de>T_f = T₀ (isothermal)

For an ideal gas, PV = nRT remains constant at fixed de>T, simplifying to de>P ∝ 1/V.

Substitute values: de>P₀V₀ = P_f(2V₀) → de>P_f = P₀/2.

Option Analysis

• A) P₀/2: Correct. Volume doubles, pressure halves per Boyle’s law for an isothermal process.
• B) 2P₀: Incorrect. Applies to isochoric heating (constant volume, pressure doubles when temperature doubles), not expansion.
• C) P₀: Incorrect. Pressure remains constant only in isobaric processes, not during volume change at fixed temperature.
• D) Undetermined: Incorrect. Initial de>P₀, V₀, T₀ completely determine the final state using the ideal gas law and isothermality.

This question tests understanding of Boyle’s law in ideal gas behavior under isothermal conditions, often encountered in CSIR NET Physics or thermodynamics topics. Doubling the volume leads to halving of pressure, consistent with de>PV = constant.