0.5 mole of an ideal gas at 300 K, expands isothermally from an initial volume of 2 L to a final volume of 6 L. Calculate : (a) work done by the gas (b) heat supplied to the gas [ Given : R = 8.31 J mol\(^{-1}\) K\(^{-1}\) ]

Q.23 0.5 mole of an ideal gas at 300 K, expands isothermally from an initial volume of 2 L to a final volume of 6 L. Calculate : (a) work done by the gas (b) heat supplied to the gas [ Given : R = 8.31 J mol\(^{-1}\) K\(^{-1}\) ]

Doubtly Asked 27 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: (a) \(W \approx 1369\) J (about 1.37 kJ) (b) \(Q = W \approx 1369\) J

(a) For an isothermal process,

\[W = nRT\ln\frac{V_f}{V_i} = 2.303\,nRT\log_{10}\frac{V_f}{V_i}\]\[W = 2.303 \times 0.5 \times 8.31 \times 300 \times \log_{10}3 = 2.303 \times 1246.5 \times 0.4771 \approx 1369.6\ \text{J}\]

(With the natural logarithm \(\ln 3 = 1.0986\), \(W = 1369.4\) J; both round to about \(1.37\times10^{3}\) J.)

(b) In an isothermal process of an ideal gas the temperature is constant, so \(\Delta U = 0\). By the first law, \(Q = \Delta U + W = W\).

Heat supplied \(Q \approx 1369\) J \(\approx 1.37\) kJ.

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?

0.5 mole of an ideal gas at 300 K, expands isothermally from an initial volume of 2 L to a final volume of 6 L. Calculate : (a) work done by the gas (b) heat supplied to the gas [ Given : R = 8.31 J mol\(^{-1}\) K\(^{-1}\) ]

1 answer

Your answer