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Write Brief Answer · Q31

Q.1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710 J and expands to 2 litres. Calculate the entropy change in expansion process.

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Step 1. With n=1n=1 mol, P=4.1P=4.1 atm and final V=2V=2 L given, the (isothermal) temperature at which this expansion occurs can be found from the ideal gas law: T=PV/(nR)=(4.1×2)/(1×0.0821)=8.2/0.0821≈99.9≈100T=PV/(nR)=(4.1\times2)/(1\times0.0821)=8.2/0.0821\approx99.9\approx100 K.

Step 2. Since the process runs at this fixed temperature (isothermal expansion, heat absorbed compensating for the work of expansion so that ΔU=0\Delta U=0 for the ideal gas), the entropy change can be found directly from ΔS=qrev/T\Delta S=q_{rev}/T. …

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