A thermally insulated vessel with nitrogen gas at $27^{\circ} \mathrm{C}$ is moving with a velocity of $100…
A thermally insulated vessel with nitrogen gas at $27^{\circ} \mathrm{C}$ is moving with a velocity of $100 \mathrm{~ms}^{-1}$. If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(assume entire loss in $\mathrm{KE}$ of the gas is given as heat to gas and $R=8.3 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ )
1.1
0.93
0.5
2.25
Solution
Let there are $n$ moles of $N_2$ gas in the cylinder.
Assuming all of $K, E$ appears in form of heat, $n\left(\frac{1}{2} M v^2\right)=\frac{f}{2} n R \Delta T$
Here, $M=28 \mathrm{~g}=28 \times 10^{-3} \mathrm{~kg}, f=5$
Increment in pressure due to change of temperature is
Also, $\Delta p=\frac{n R \Delta T}{V}$
So, $\frac{\Delta p}{p}=\frac{\left(\frac{n R \Delta T}{V}\right)}{\left(\frac{n R T}{V}\right)}=\frac{n R \Delta T}{n R T} \Rightarrow \frac{\Delta p}{p}=\frac{n M v^2}{f n R T}=\frac{M v^2}{f R T}$
So, percentage change in pressure is,
$
\therefore \frac{\Delta p}{p} \times 100=\frac{28 \times 10^{-3} \times 100 \times 100}{5 \times 8.3 \times 300} \times 100=2.25 \%
$