Two moles of an ideal monoatomic gas at 27°C occupies a volume of V. If the gas is expanded adiabatically to the volume 2V, then the work done by the gas in S.I. system is $\frac{1000 x}{3} [4^{1 / 3} = 1 .6]$ . The value of $x$ is………..(Round off to nearest Integer).

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answer is 8.

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Detailed Solution

$In adiabatic process,$

${TV}^{γ - 1} = cons tant$

$\Rightarrow 300 \times V^{\frac{5}{3} - 1} = T_{2} (2 V)^{\frac{5}{3} - 1}$

$\Rightarrow T_{2} = \frac{300}{2^{\frac{2}{3}}} = \frac{300}{4^{\frac{1}{3}}} = \frac{300}{1.6}$

Work done in adiabatic process is,

$W = \frac{nR (T_{1} - T_{2})}{γ - 1}$

$\Rightarrow W = \frac{2 \times 8.314 \times (300 - 187.5) \times 3}{2} = \frac{1000 x}{3}$

On solving, We get $x = 8.4$

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