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Thermodynamics

JEE Main level, numerical answer, medium

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Question 2 of 3

In an isothermal reversible expansion, $2\,\text{moles}$ of an ideal gas expand from $V_1 = 1\,\text{L}$ to $V_2 = 10\,\text{L}$ at $300\,\text{K}$ . Work done is (in J, take $R=8.314$ , $\ln 10 = 2.303$ ):

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Answer: 11488

W = 2.303 n R T \log(V_2/V_1) = 2.303 \times 2 \times 8.314 \times 300 \times 1 = 11488\,\text{J}

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Chemistry practice question on Thermodynamics, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

In an isothermal reversible expansion, 2 moles2\,\text{moles}2moles of an ideal gas expand from V1=1 LV_1 = 1\,\text{L}V1​=1L to V2=10 LV_2 = 10\,\text{L}V2​=10L at 300 K300\,\text{K}300K. Work done is (in J, take R=8.314R=8.314R=8.314, ln⁡10=2.303\ln 10 = 2.303ln10=2.303):

JEE Main question paper PDF

More Chemistry practice

In an isothermal reversible expansion, $2\,\text{moles}$ of an ideal gas expand from $V_1 = 1\,\text{L}$ to $V_2 = 10\,\text{L}$ at $300\,\text{K}$ . Work done is (in J, take $R=8.314$ , $\ln 10 = 2.303$ ):