MHT CET · Chemistry · Thermodynamics (C)
What is the work done when a gas is compressed from \(2.5 \times\) \(10^{-2} \mathrm{~m}^3\) to \(1.3 \times 10^{-2} \mathrm{~m}^3\) at constant external pressure of 4.05 bar?
- A \(4050 \mathrm{~J}\)
- B \(4400 \mathrm{~J}\)
- C \(4200 \mathrm{~J}\)
- D \(4860 \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(4860 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{W}=-\operatorname{Pext}\left[\mathrm{V}_2-\mathrm{V}_1\right] \\
& =-4.05\left[1.3 \times 10^{-2}-2.5 \times 10^{-2}\right] \\
& =4.05 \times 1.2 \times 10^{-2}=4.86 \times 10^{-2} \mathrm{bar}-\mathrm{m}^3 \\
& \left(1 \mathrm{bar}=10^5 \mathrm{~Pa}, 1 \mathrm{~Pa}-\mathrm{m}^3=1 \mathrm{~J}\right) \\
& \mathrm{W}=4.86 \times 10^{-2} \times 10^5 \mathrm{~Pa}-\mathrm{m}^2 \\
& =4860 \mathrm{~J}
\end{aligned}
\)
\begin{aligned}
& \mathrm{W}=-\operatorname{Pext}\left[\mathrm{V}_2-\mathrm{V}_1\right] \\
& =-4.05\left[1.3 \times 10^{-2}-2.5 \times 10^{-2}\right] \\
& =4.05 \times 1.2 \times 10^{-2}=4.86 \times 10^{-2} \mathrm{bar}-\mathrm{m}^3 \\
& \left(1 \mathrm{bar}=10^5 \mathrm{~Pa}, 1 \mathrm{~Pa}-\mathrm{m}^3=1 \mathrm{~J}\right) \\
& \mathrm{W}=4.86 \times 10^{-2} \times 10^5 \mathrm{~Pa}-\mathrm{m}^2 \\
& =4860 \mathrm{~J}
\end{aligned}
\)
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