JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Chemical Thermodynamics

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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

lo g 2 = 0.3010

)

Answer: 8630

Solution

For isothermal reversible expansion of an ideal gas, the work done is:

w = - n R T ln \frac{V _{2}}{V _{1}}

Given:

n = 5 mol
R = 8.3 J K⁻¹ mol⁻¹
T = 300 K
V₁ = 10 L, V₂ = 20 L

Substituting:

w = - 5 \times 8.3 \times 300 \times ln \frac{20}{10}

w = - 5 \times 8.3 \times 300 \times ln 2

w = - 12450 \times 2.303 \times lo g 2

w = - 12450 \times 2.303 \times 0.3010

w = - 8630.38 J

Magnitude of maximum work = 8630 J (nearest integer)

Answer: 8630

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer

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