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