Q35 / 150 · MHT CET 2021 (20 Sep Shift 2) · Chemistry

An ideal gas on isothermal reversible compression from $10 \mathrm{~L}$ to $5 \mathrm{~L}$ performs $1730 \mathrm{~J}$ of work at $300 \mathrm{~K}$. Calculate number of moles of gas involved in compression? $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

  • A
    2.5
  • B
    2
  • Correct answer
    C
    1
  • D
    3

Solution

$$ \begin{aligned} & \mathrm{V}_1=10 \mathrm{~L}, \mathrm{~V}_2=5 \mathrm{~L}, \mathrm{~W}=1730 \mathrm{~J}, \mathrm{~T}=300 \mathrm{~K}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{n} \\ & =? \\ & \mathrm{~W}_{\max }=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & \therefore 1730 \mathrm{~J}=-2.303 \mathrm{n} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \times \log _{10}\left(\frac{5}{10}\right) \\ & \therefore \mathrm{n}=\frac{1730 \mathrm{~J}}{-2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{mo}^{-1} \times 300 \mathrm{~K} \times \log _{10}(0.5)} \\ & \therefore \mathrm{n}=\frac{1730}{-5744.14 \times(-0.3010)} \mathrm{mol} \\ & \therefore \mathrm{n}=\frac{1730}{1729} \approx 1 \mathrm{~mol} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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