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If the ratio \mathrm{C_p / C_v=\gamma}, the change in internal energy of the mass of a gas, when the volume changes from V to 2 V at constant pressureP is

Option: 1

\mathrm{\frac{R}{(\gamma-1)}}


Option: 2

PV


Option: 3

\mathrm{\frac{P V}{(\gamma-1)}}


Option: 4

\mathrm{\frac{\gamma P V}{(\gamma-1)}}


Answers (1)

best_answer

Let ?T be the increase in temperature when the volume of the gas is changed by ?V at
constant pressure. The change in internal energy of n moles of a gas is given by

\mathrm{\Delta U=n C_v \Delta T}           .....[i]

\mathrm{\text { We know that } C_p-C_v=R}

\mathrm{\text { Or } \quad \frac{C_p}{C_v}=1+\frac{R}{C_v} \text {. But } \frac{C_p}{C_v}=\gamma}

\mathrm{\text { Therefore, } \gamma=1+\frac{R}{C^2} \text {, which gives }}

\mathrm{C_v=\frac{R}{(\gamma-1)}}                 ....[ii]

Also PV = n RT. At constant pressure, when volume changes by ?V, the change in
temperature ?T is given by

\mathrm{\begin{aligned} & P \Delta V=n R \Delta T \\ & \text { Or } \Delta T=\frac{P \Delta V}{n R}=\frac{P V}{n R} \\ & (\because \Delta V=2 V-V=V) \end{aligned}}               ....[iii]

Using (ii) and (iii) in (i) we have

\mathrm{\Delta U=n \times \frac{R}{(\gamma-1)} \times \frac{P V}{n R}=\frac{P V}{(\gamma-1)}}

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jitender.kumar

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