MHT CET · Chemistry · Thermodynamics (C)
When 2 moles of an ideal gas are expanded isothermally from a volume of \(12.5 \mathrm{~L}\) to 15.0 L against constant external pressure of \(760 \mathrm{~mm} \mathrm{Hg}\). Calculate the amount of work done in joule?
- A \(-253 \cdot 25 \mathrm{~J}\)
- B \(-190 \cdot 0 \mathrm{~J}\)
- C \(-1924 \cdot 0 \mathrm{~J}\)
- D \(-25 \cdot 325 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(-253 \cdot 25 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{n} =2 \mathrm{~mol}, \quad \mathrm{~V}_{1}=12.5 \mathrm{~L}=12.5 \times 10^{-3} \mathrm{~m}^{3},\) \(\mathrm{~V}_{2}=15.0 \mathrm{~L}=15.0 \times 10^{-3} \mathrm{~m}^{3}, \)
\( \mathrm{P} =760 \mathrm{~mm} \text { of } \mathrm{Hg}=1.013 \times 10^{5} \mathrm{Nm}^{-2} \)
\( \mathrm{~W} =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)=-1.013 \times 10^{5}\)\((15.0-12.5) 10^{-3} \)
\( =-2.5325 \times 10^{5} \times 10^{-3} \)
\( \therefore \quad \mathrm{W} =-253.25 \mathrm{~J}\)
\( \mathrm{P} =760 \mathrm{~mm} \text { of } \mathrm{Hg}=1.013 \times 10^{5} \mathrm{Nm}^{-2} \)
\( \mathrm{~W} =-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)=-1.013 \times 10^{5}\)\((15.0-12.5) 10^{-3} \)
\( =-2.5325 \times 10^{5} \times 10^{-3} \)
\( \therefore \quad \mathrm{W} =-253.25 \mathrm{~J}\)
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