An ideal gas undergoes a process in which $P V^{- 2} = constant$ , where V is the volume occupied by the gas initially at pressure P. At the end of the process, rms speed of gas molecules has became $2^{1 / 2}$ times of its initial value. Find the value of $C_{v}$ so that energy transferred by the heat to the gas is ‘2’ times of the initial energy.

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$\frac{5 R}{3}$

$\frac{3 R}{5}$

$3 R$

$\frac{R}{3}$

answer is D.

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Detailed Solution

$P V^{- 2} = C o n s \tan t$

Molar heat capacity in the process
$P V^{x} = C o n s \tan t$ , is
$C = \frac{R}{r - 1} + \frac{R}{(1 - x)} = C_{v} + \frac{R}{1 - (- 2)} = C_{v} + \frac{R}{3}$
At the end of the process $V_{r m s}$ is $2^{1 / 2}$ times. Temperature has become ‘2’ times. $(V_{r m s} \propto T^{1 / 2})$
$Δ Q = n C Δ T = n C T (a - 1) = n [C_{v} + \frac{R}{3}] (2 T - T)$
But given $Δ Q = 2 U_{i}$

$\Rightarrow Δ Q = 2 n C_{V} T$
Or $n [C_{V} + \frac{R}{3}] (T) = 2 n C_{v} T$

Or $C_{V} = \frac{R}{3}$

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