Q.18 One mole of an ideal gas expands isothermally and reversibly to double its volume.

The temperature of the system is 300.00 K.

Problem Solution

For one mole of an ideal gas undergoing reversible isothermal expansion to double its volume, the work done by the system follows W = nRT ln(V₂/V₁), where n = 1, V₂/V₁ = 2, so W = RT ln 2.

Given W = 1728.85 J and R = 8.314 J K⁻¹ mol⁻¹, rearrange to T = W / (R ln 2).

With ln 2 ≈ 0.693147, T = 1728.85 / (8.314 × 0.693147) ≈ 300.00 K.

SEO Article Content

One mole of an ideal gas expands isothermally and reversibly to double its volume, performing 1728.85 J of expansion work with R = 8.314 J K⁻¹ mol⁻¹. This isothermal reversible expansion ideal gas work double volume temperature calculation is key for CSIR NET Life Sciences thermodynamics.

Core Formula Derivation

Work done by the gas in reversible isothermal expansion derives from dW = P dV and ideal gas law P = nRT/V, yielding W = nRT ln(V₂/V₁).

For n=1 and V₂/V₁ = 2, simplifies to W = RT ln 2.

Base-10 form is W = 2.303 RT log₁₀ 2, but natural log suits direct computation.

Step-by-Step Calculation

• Substitute values: ln 2 = 0.693147, so denominator R ln 2 = 8.314 × 0.693147 = 5.7628
• Then T = 1728.85 / 5.7628 = 300.0004 K, rounded to 300.00 K
• This matches standard examples where work scales with T for fixed volume ratio

Exam Relevance

• No options provided, but numerical fits GATE/CSIR NET style (e.g., similar to neon expansion problems)
• Internal energy change ΔU=0 for ideal gas isothermal process, so q = -w = 1728.85 J (heat absorbed)
• Practice verifies precision: slight W variation confirms exact 300 K design