The work done during reversible isothermal expansion of one mole of hydrogen gas at 25^ from a pressure of 20 atm to 10 atm is:(Given R = 2.0 cal K^-1 - Chemistry MCQ NEET (UG) 2024 2026 | ExamTest.live

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The work done during reversible isothermal expansion of one mole of hydrogen gas at 2 5^{\circ} C from a pressure of 20 atm to 10 atm is:(Given R = 2.0 cal K^{- 1} mol^{- 1}, ln 2 \approx 0.693)

Solution & Step-by-step Explanation

For reversible isothermal process: W = - n RT ln (\frac{V _{2}}{V _{1}}) = - n RT ln (\frac{P _{1}}{P _{2}}) Given: n = 1 mol, R = 2 cal K^{- 1} mol^{- 1}, T = 25 + 273 = 298 K, P_{1} = 20 atm, P_{2} = 10 atm . W = - (1) \times (2) \times (298) \times ln (\frac{20}{10}) W = - 596 \times ln 2 \approx - 596 \times 0.693 W \approx - 413.03 cal The closest option is - 413.14 calories .

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The work done during reversible isothermal expansion of one mole of hydrogen gas at 2 5^{\circ} C from a pressure of 20 atm to 10 atm is:(Given R = 2.0 cal K^{- 1} mol^{- 1}, ln 2 \approx 0.693)

- 413.14 calories

413.14 calories

100 calories

0 calorie

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