One mole of {O_2} gas having a volume equal t | vedclass

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(D) For an isothermal process,the work done $W$ is given by the formula: $W = -\mu RT \ln\left(\frac{V_2}{V_1}ight)$.Given: $\mu = 1 	ext{ mole}$,$

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$-1572.5$

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(D) For an isothermal process,the work done $W$ is given by the formula: $W = -\mu RT \ln\left(\frac{V_2}{V_1}ight)$.Given: $\mu = 1 	ext{ mole}$,$T = 273 	ext{ K}$,$V_1 = 22.4 	ext{ litres}$,$V_2 = 11.2 	ext{ litres}$,and $R = 8.31 	ext{ J/mol K}$.Substituting the values:$W = -1 	imes 8.31 	imes 273 	imes \ln\left(\frac{11.2}{22.4}ight)$.$W = -8.31 	imes 273 	imes \ln(0.5)$.Since $\ln(0.5) = -\ln(2) \approx -0.693$,$W = -8.31 	imes 273 	imes (-0.693) \approx 1572.5 	ext{ J}$.Note: The work done on the gas is positive,but the work done by the gas is negative. Given the options,the value $-1572.5 	ext{ J}$ represents the work done by the gas during compression.

One mole of ${O_2}$ gas having a volume equal to $22.4 \text{ litres}$ at $0^{\circ}C$ and $1 \text{ atmospheric pressure}$ is compressed isothermally so that its volume reduces to $11.2 \text{ litres}$. The work done in this process is ...... $J$.

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