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NEET CHEMISTRYThermodynamicsMedium

Question

Work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from a pressure of 20 atmospheres to a pressure of 10 atmospheres is: (Given: R = 2.0 cal K–1 mol–1)

A

–413.14 calories

B

413.14 calories

C

100 calories

D

0 calorie

Step-by-Step Solution

The work done (ww) during the reversible isothermal expansion of an ideal gas is given by the formula: w=2.303nRTlog10(V2V1)w = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) Since P1V1=P2V2P_1V_1 = P_2V_2 for an isothermal process, we can also write this as: w=2.303nRTlog10(P1P2)w = -2.303 nRT \log_{10}\left(\frac{P_1}{P_2}\right)

Given the values: Number of moles, n=1n = 1 mol Universal gas constant, R=2.0 cal K1mol1R = 2.0 \text{ cal K}^{-1} \text{mol}^{-1} Temperature, T=25C=25+273=298 KT = 25^\circ\text{C} = 25 + 273 = 298 \text{ K} Initial pressure, P1=20 atmP_1 = 20 \text{ atm} Final pressure, P2=10 atmP_2 = 10 \text{ atm}

Substituting these values into the formula: w=2.303×1×2.0×298×log10(2010)w = -2.303 \times 1 \times 2.0 \times 298 \times \log_{10}\left(\frac{20}{10}\right) w=2.303×596×log10(2)w = -2.303 \times 596 \times \log_{10}(2) w=2.303×596×0.3010w = -2.303 \times 596 \times 0.3010 w413.14 caloriesw \approx -413.14 \text{ calories}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsduringreversibleisothermalexpansionhydrogen

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