Question

A thermally insulated vessel with nitrogen gas at $27^{\circ} C$ is moving with a velocity of $100\, ms ^{-1} .$ If the vessel is stopped suddenly, then the percentage change in the pressure of the gas is nearly
(assume entire loss in KE of the gas is given as heat to gas and $R=8.3\, Jmol ^{-1} K ^{-1}$ )

• A. 1.1
• B. 0.93
• C. 0.5
• D. 2.25

Answer 1

Answer: (D)

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, $M=28\, g =28 \times 10^{-3} kg ,\, f=5$
Increment in pressure due to change of temperature is
Also, $\Delta p=\frac{n R \Delta T}{V}$
So, $\frac{\Delta p}{p}=\frac{\left(\frac{n R \wedge T}{V}\right)}{\left(\frac{n R T}{V}\right)}=\frac{n R \Delta T}{n R T}$
$\Rightarrow \frac{\Delta p}{p}=\frac{n M v^{2}}{f n R T}=\frac{M v^{2}}{f R T}$
So, percentage change in pressure is,
$\therefore \frac{\Delta p}{p} \times 100=\frac{28 \times 10^{-3} \times 100 \times 100}{5 \times 8.3 \times 300} \times 100=2.25 \%$