An ideal gas expands isothermally and reversibly from $10 \mathrm{~m}^{3}$ to $20 \mathrm{~m}^{3}$ at $300…

An ideal gas expands isothermally and reversibly from $10 \mathrm{~m}^{3}$ to $20 \mathrm{~m}^{3}$ at $300 \mathrm{~K}$ performing $5 \cdot 187 \mathrm{~kJ}$ of work on surrrounding. Calculate number of moles of gas undergoing expansion? $\left(\mathrm{R}=8 \cdot 314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.$ )
  1. 1.5
  2. 2
  3. 3
  4. 1

Solution

$\mathrm{V}_{1}=10 \mathrm{~m}^{3}, \mathrm{~V}_{2}=20 \mathrm{~m}^{3}, \mathrm{~T}=300 \mathrm{~K}$ $\mathrm{W}_{\max }=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}, \mathrm{n}=?$ $\mathrm{W}_{\max }=-2.303 \mathrm{~nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$ $\therefore-5187=-2.303 \times n \times 8.314 \times 300 \log _{10} \frac{20}{10}$ $\therefore \mathrm{n}=\frac{5187}{2.303 \times 8.314 \times 300 \times \log _{10} 2}=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}$ $\therefore \mathrm{n}=3 \mathrm{~mol}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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