Q.420. First Law of Thermodynamics.

Question 420.
Calculate the work, heat, delta U, delta H for the reversible isothermal expansion of 2 moles of ideal gas from 16 L to 95 L at 0 degrees celcius.

Answer 420.
For isothermal expansion, temperature being constant,
Change in internal energy, ΔU = 0.
Work done by the ideal gas in reversible isothermal process, W
= nRT ln(V2/v1)
= 2 * 8.31 * 273 * ln(95/16) J
= 8082 J.
Heat absorbed by the gas, Q
= work done by the gas
= 8082 J
= 8082 * 0.239 calories
= 1932 calories.
ΔH = ∫(T1 to T2) Cp dT = 0 as T1 = T2.

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