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(D) Let $n$ be the number of moles of $N_2$ gas. The kinetic energy of the gas is $K.E. = n \left( \frac{1}{2} M v^2 \right)$,where $M$ is the molar m

Vedclass · Accepted Answer

$2.25$

Vedclass · Answer

(D) Let $n$ be the number of moles of $N_2$ gas. The kinetic energy of the gas is $K.E. = n \left( \frac{1}{2} M v^2 ight)$,where $M$ is the molar mass of $N_2$ $(28 	imes 10^{-3} \ kg/mol)$.Since the vessel is insulated,the loss in kinetic energy is converted into internal energy (heat) of the gas: $n \left( \frac{1}{2} M v^2 ight) = n C_v \Delta T$.For a diatomic gas like $N_2$,the degrees of freedom $f = 5$,so $C_v = \frac{f}{2} R = \frac{5}{2} R$.Thus,$\frac{1}{2} M v^2 = \frac{5}{2} R \Delta T \Rightarrow \Delta T = \frac{M v^2}{5 R}$.From the ideal gas law $PV = nRT$,the change in pressure $\Delta P$ for a constant volume process is $\Delta P = \frac{nR \Delta T}{V} = \frac{P \Delta T}{T}$.The percentage change in pressure is $\frac{\Delta P}{P} 	imes 100 = \frac{\Delta T}{T} 	imes 100$.Substituting $\Delta T = \frac{M v^2}{5 R}$ and $T = 300 \ K$:$\frac{\Delta P}{P} 	imes 100 = \frac{M v^2}{5 R T} 	imes 100 = \frac{28 	imes 10^{-3} 	imes 100^2}{5 	imes 8.3 	imes 300} 	imes 100 = \frac{280}{12450} 	imes 100 \approx 2.25 \%$.

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