If the ratio $\frac{C_{P}}{C_{V}} = γ$ , the change in internal energy of the mass of a gas, when the volume changes from V to 2V at constant pressure ‘P’ is

\frac{R}{γ - 1}

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P V

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\frac{P V}{(γ - 1)}

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\frac{γ P V}{(γ - 1)}

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Solution

The correct option is C $\frac{P V}{(γ - 1)}$
We know that $Δ U = n C_{V} Δ T$ and $C_{V} = \frac{R}{r - 1}$
At constant pressure, volume changes by $D e l t a$ V so
$p Δ V = n R Δ T$
$\Rightarrow Δ T = \frac{P (Δ v)}{n R} = \frac{p v}{n R} [∵ Δ v = 2 v - v = v]$
$∴ Δ v = n \frac{R}{r - 1} (\frac{p v}{n R}) = \frac{P V}{(r - 1)}$

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