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Question 56 of 68

Q.Derive an expression for the work done during an isothermal process.
104 J of work is done on certain volume of a gas. If the gas releases 125 kJ of heat, calculate the change in internal energy of the gas.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 4mImportance★★★★★
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Isothermal work from PV=const, then applying the first law with signs for the given data.

Work done in an isothermal process: At constant temperature, PV=nRT=constPV=nRT=\text{const}, so P=nRT/VP = nRT/V. Work done by the gas as it expands from V1V_1 to V2V_2:

W=∫V1V2P dV=∫V1V2nRTV dV=nRTln⁡V2V1=2.303 nRTlog⁡10V2V1W = \int_{V_1}^{V_2}P\,dV = \int_{V_1}^{V_2}\frac{nRT}{V}\,dV = nRT\ln\frac{V_2}{V_1} = 2.303\,nRT\log_{10}\frac{V_2}{V_1}

Numerical part: Work is done on the gas, =104= 104 J, so work done by the gas is W=−104W = -104 J. The gas releases 125 kJ of heat, so heat supplied to the gas is Q=−125000Q = -125000 J.

By the first law, Q=ΔU+WQ = \Delta U + W: …

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