Question:
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, \(27^\circ C\). The work done on the gas will be :
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, \(27^\circ C\). The work done on the gas will be :
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Isothermal work on n moles: \(W = nRT\ln(P_2/P_1)\) with T = 300 K and \(P_2/P_1 = 2\).
Updated On: Nov 11, 2026
- \(300R\)
- \(300R\,\ln6\)
- \(300R\,\ln2\)
- \(300R\,\ln7\)
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The Correct Option is C
Solution and Explanation
For an isothermal process, \(P_1V_1 = P_2V_2\); since \(P_2 = 2P_1\), \(V_2 = V_1/2\).
Work done on the gas: \(W = nRT\ln\frac{V_1}{V_2} = nRT\ln\frac{P_2}{P_1}\).
With n = 1, T = 27 + 273 = 300 K and \(P_2/P_1 = 2\): \(W = 300R\ln2\).
Answer: (3) \(300R\,\ln2\)
Work done on the gas: \(W = nRT\ln\frac{V_1}{V_2} = nRT\ln\frac{P_2}{P_1}\).
With n = 1, T = 27 + 273 = 300 K and \(P_2/P_1 = 2\): \(W = 300R\ln2\).
Answer: (3) \(300R\,\ln2\)
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