One mole of $O_{2}$ gas having a volume equal to 22.4 litres at $0^{0} C$ and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is :

1672.5J

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1728J

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-1728J

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-1572.5J

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Solution

The correct option is D -1572.5J
Work done in adiabatic process,

W = $μ R T$ $l o g_{e} \frac{V_{e}}{V_{r}}$
W = $1 \times 8.314 \times 273 \times l o g_{e} (\frac{11.2}{22.4})$
W = $8.314 \times 273 \times l o g_{e} (\frac{1}{2})$
W = $8.314 \times 273 \times [l o g_{e} (1) - l o g_{e} (2)]$
W = $8.314 \times 273 \times (0 - 0.69)$
W = -1566.1 J

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One mole of O2 gas having a volume equal to 22.4 litres at 00C and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is :

The correct option is D -1572.5JWork done in adiabatic process,W = μRT logeVeVrW = 1×8.314×273×loge(11.222.4)W = 8.314×273×loge(12)W = 8.314×273×[loge(1)−loge(2)]W = 8.314×273×(0−0.69)W = -1566.1 J

One mole of $O_{2}$ gas having a volume equal to 22.4 litres at $0^{0} C$ and 1 atmospheric pressure is compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is :