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  • P − Vplots for two gases during adiabatic processes are shown in Fig. Plots 1 and 2 should correspond respectively to <img alt="" src="https://cdn.entrance360.com/media/uploads/2023

P − Vplots for two gases during adiabatic processes are shown in Fig. Plots 1 and 2 should correspond respectively to

Option: 1

\mathrm{He} \text { and } \mathrm{O}_2


Option: 2

\mathrm{O}_2 \text { and } \mathrm{He}


Option: 3

He and Ar


Option: 4

\mathrm{O}_2 \text { and } \mathrm{N}_2


Answers (1)

best_answer

For an adiabatic process, \mathrm{P V^\gamma}= constant, Differentiating, we have

\mathrm{\gamma P V^{\gamma-1}+\frac{d P}{d V} V^\gamma=0 \text { or } \frac{d P}{d V}=-\frac{\gamma P}{V}}

Since at any instant PV = constant, \mathrm{\frac{d P}{d V} \propto \gamma}, i.e. the slope of P − V curve is proportional to \mathrm{\gamma}. Now, for a diatomic gas, \mathrm{\gamma(=7 / 5)} is than that of for a monoatomic gas for which \mathrm{\gamma=5 / 3}. Therefore, the slope of the P − V curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas and curve 2 to monoatomic gas

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Gaurav

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