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One mole of $\mathrm{O}_{2}$ gas having a volume equal to 22.4 Litres at $0^{\circ} \mathrm{C}$ and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is-
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$-1572.5 \mathrm{~J}$
Work done in adiabatic process, \(\mathrm{W}=\mu \mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)\)
\(\mathrm{W}=1 \times 8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{11.2}{22.4}\right)\)
\(\mathrm{W}=8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{1}{2}\right)\)
\(\mathrm{W}=8.314 \times 273 \times\left[\log _{\mathrm{e}}(1)-\log _{\mathrm{e}}(2)\right]\)
\(\mathrm{W}=8.314 \times 273 \times(0-0.69)\)
\(\mathrm{W}=-1566.1 \mathrm{~J}\)
\(\mathrm{W}=1 \times 8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{11.2}{22.4}\right)\)
\(\mathrm{W}=8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{1}{2}\right)\)
\(\mathrm{W}=8.314 \times 273 \times\left[\log _{\mathrm{e}}(1)-\log _{\mathrm{e}}(2)\right]\)
\(\mathrm{W}=8.314 \times 273 \times(0-0.69)\)
\(\mathrm{W}=-1566.1 \mathrm{~J}\)
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