MHT CET · Chemistry · Thermodynamics (C)
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? \(\left(\mathrm{R}=8 \cdot 314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.\) )
- A 1.5
- B 2
- C 3
- D 1
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
\(\mathrm{V}_{1}=10 \mathrm{~m}^{3}, \mathrm{~V}_{2}=20 \mathrm{~m}^{3}, \mathrm{~T}=300 \mathrm{~K}\)
\(\mathrm{W}_{\max }=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}, \mathrm{n}=?\)
\(\mathrm{W}_{\max }=-2.303 \mathrm{~nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)
\(\therefore-5187=-2.303 \times n \times 8.314 \times 300 \log _{10} \frac{20}{10}\)
\(\therefore \mathrm{n}=\frac{5187}{2.303 \times 8.314 \times 300 \times \log _{10} 2}=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}\)
\(\therefore \mathrm{n}=3 \mathrm{~mol}\)
\(\mathrm{W}_{\max }=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}, \mathrm{n}=?\)
\(\mathrm{W}_{\max }=-2.303 \mathrm{~nRT} \log _{10} \frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)
\(\therefore-5187=-2.303 \times n \times 8.314 \times 300 \log _{10} \frac{20}{10}\)
\(\therefore \mathrm{n}=\frac{5187}{2.303 \times 8.314 \times 300 \times \log _{10} 2}=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}\)
\(\therefore \mathrm{n}=3 \mathrm{~mol}\)
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