MHT CET · Chemistry · Thermodynamics (C)
What is the work done when 2 mole of an ideal gas are expanded isothermally and reversibly from \(5 \mathrm{~m}^{3}\) to \(10 \mathrm{~m}^{3}\) at \(300 \mathrm{~K} ?\left(\mathrm{R}=8 \cdot 314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(-34.58 \mathrm{~kJ}\)
- B \(3 .458 \mathrm{~kJ}\)
- C \(-1.728 \mathrm{~kJ}\)
- D \(-3 \cdot 458 \mathrm{k}\)
Answer & Solution
Correct Answer
(D) \(-3 \cdot 458 \mathrm{k}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{n}=2\) mole, \(\quad V_{1}=5 \mathrm{~m}^{3}, V_{2}=10 \mathrm{~m}^{3}\)
\(\mathrm{T}=300 \mathrm{~K}, \quad \mathrm{R}=8.314 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}\)
\(\mathrm{W}_{\max }=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)
\(\therefore W_{\max }=-2.303 \times 2 \times 8.314 \times 300 \log _{10} \frac{10}{5}\)
\(=-3457.97 \mathrm{~J}\)
\(\therefore \mathrm{W}_{\max }=-3.458 \mathrm{~kJ}\)
\(\mathrm{T}=300 \mathrm{~K}, \quad \mathrm{R}=8.314 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}\)
\(\mathrm{W}_{\max }=-2.303 \mathrm{nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)
\(\therefore W_{\max }=-2.303 \times 2 \times 8.314 \times 300 \log _{10} \frac{10}{5}\)
\(=-3457.97 \mathrm{~J}\)
\(\therefore \mathrm{W}_{\max }=-3.458 \mathrm{~kJ}\)
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