Q24 / 90 · JEE Main 2022 (27 Jun Shift 2) · Chemistry

When $5$ moles of $He$ gas expand isothermally and reversibly at $300 K$ from $10$ litre to $20$ litre, the magnitude of the maximum work obtained is $J$ . [nearest integer ] (Given : $R = 8.3 J K^{- 1} {mol}^{- 1}$ and $\log 2 = 0.3010$ )

Correct answer: 8630

Solution

Work done in isothermal reversible process is calculated by the equation

$W = - 2.303 nRTlog (\frac{V_{2}}{V_{1}}) n = Number of moles T = Temperature = 300 K V_{2} = Final volume = 20 litres V_{1} = Initial volume = 10 litres$

W $= - 2.303 \times 5 \times 8.3 \times 300 \log (\frac{20}{10})$

$= - 8630.38 J$

Hence, the correct answer is 8630.

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When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is J. [nearest integer ] (Given : R=8.3 J K-1 mol-1 and log2=0.3010)

Solution

Relevant Questions

When $5$ moles of $He$ gas expand isothermally and reversibly at $300 K$ from $10$ litre to $20$ litre, the magnitude of the maximum work obtained is $J$ . [nearest integer ] (Given : $R = 8.3 J K^{- 1} {mol}^{- 1}$ and $\log 2 = 0.3010$ )