MHT CET · Chemistry · Thermodynamics (C)
Two moles of an ideal gas is expanded isothermally from a volume of \(300 \mathrm{~cm}^3\) to 2.5 \(\mathrm{dm}^3\) at 298 K against a constant pressure at 1.9 bar. Calculate the work done in joules.
- A \(-418\text J\)
- B \(-565 \mathrm{~J}\)
- C \(-918\text J\)
- D \(-950 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(-418\text J\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{V}_1=300 \mathrm{~cm}^3=0.3 \mathrm{dm}^3,\left(\because 1 \mathrm{dm}^3=1.000 \mathrm{~cm}^3\right) \\ & \mathrm{V}_2=2.5 \mathrm{dm}^3\end{aligned}\)
\(\begin{aligned} \mathrm{P}_{\mathrm{ext}} & =1.9 \mathrm{bar} \\ \mathrm{W} & =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\ & =-1.9(2.5-0.3) \\ & =-4.18 \mathrm{dm}^3 \mathrm{bar} \times 100 \frac{\mathrm{~J}}{\mathrm{dm}^3 \mathrm{bar}} \\ \therefore \quad \mathrm{W} & =-418 \mathrm{~J}\end{aligned}\)
\(\begin{aligned} \mathrm{P}_{\mathrm{ext}} & =1.9 \mathrm{bar} \\ \mathrm{W} & =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\ & =-1.9(2.5-0.3) \\ & =-4.18 \mathrm{dm}^3 \mathrm{bar} \times 100 \frac{\mathrm{~J}}{\mathrm{dm}^3 \mathrm{bar}} \\ \therefore \quad \mathrm{W} & =-418 \mathrm{~J}\end{aligned}\)
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