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 surrounding, calculate number of moles of gas used?
- A 1
- B 3
- C 2
- D 1.5
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
(A)
\(\begin{array}{l}
V_{1}=10 \mathrm{~m}^{3}=10 \times 10^{3} \mathrm{dm}^{3} \\
V_{2}=20 \mathrm{~m}^{3}=20 \times 10^{3} \mathrm{dm}^{3}
\end{array}\)
\(T=300 K, W=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}\)
\(W=-2.303 \mathrm{nRT} \log \frac{V_{2}}{V_{1}}\)
\(\therefore-5187=-2.303 \times n \times 8.314 \times 300\) \(\times \log _{10} \frac{20 \times 10^{3}}{10 \times 10^{3}}\)
\(\therefore n=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.301}\)
\(\therefore n=3\) moles
\(\begin{array}{l}
V_{1}=10 \mathrm{~m}^{3}=10 \times 10^{3} \mathrm{dm}^{3} \\
V_{2}=20 \mathrm{~m}^{3}=20 \times 10^{3} \mathrm{dm}^{3}
\end{array}\)
\(T=300 K, W=-5.187 \mathrm{~kJ}=-5187 \mathrm{~J}\)
\(W=-2.303 \mathrm{nRT} \log \frac{V_{2}}{V_{1}}\)
\(\therefore-5187=-2.303 \times n \times 8.314 \times 300\) \(\times \log _{10} \frac{20 \times 10^{3}}{10 \times 10^{3}}\)
\(\therefore n=\frac{5187}{2.303 \times 8.314 \times 300 \times 0.301}\)
\(\therefore n=3\) moles
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