MHT CET · Chemistry · Thermodynamics (C)
When certain volume of gas expands against a constant external pressure of \(2.40 \times 10^5 \mathrm{~Pa}\) at \(300 \mathrm{~K}\) to \(2.2 \times 10^{-3} \mathrm{~m}^3\). If the work obtained is \(-0.048 \mathrm{~kJ}\). What is the initial volume of the gas?
- A \(2 \times 10^{-3} \mathrm{~m}^3\)
- B \(4.5 \times 10^{-2} \mathrm{~m}^3\)
- C \(1.5 \times 10^{-3} \mathrm{~m}^3\)
- D \(2.8 \times 10^{-2} \mathrm{~m}^3\)
Answer & Solution
Correct Answer
(A) \(2 \times 10^{-3} \mathrm{~m}^3\)
Step-by-step Solution
Detailed explanation
\(\mathrm{W}=-\mathrm{P}_{\mathrm{ext}}\left[\mathrm{V}_2-\mathrm{V}_1\right] \)
\( -0.048 \times 1000 \mathrm{~J}=-2.4 \times 10^5 \mathrm{~Pa}[2.2 \times\) \(10^{-3}-\mathrm{V}_1] \mathrm{m}^3 \)
\( \frac{48}{2.8} \times 10^{-5}=2.2 \times 10^{-3}-\mathrm{V}_1 \)
\( \mathrm{~V}_1=2.2 \times 10^{-3}-0.2 \times 10^{-3} \)
\( \mathrm{~V}_1=2 \times 10^{-3} \mathrm{~m}^3\)
\( -0.048 \times 1000 \mathrm{~J}=-2.4 \times 10^5 \mathrm{~Pa}[2.2 \times\) \(10^{-3}-\mathrm{V}_1] \mathrm{m}^3 \)
\( \frac{48}{2.8} \times 10^{-5}=2.2 \times 10^{-3}-\mathrm{V}_1 \)
\( \mathrm{~V}_1=2.2 \times 10^{-3}-0.2 \times 10^{-3} \)
\( \mathrm{~V}_1=2 \times 10^{-3} \mathrm{~m}^3\)
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