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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

A thermally insulated vessel with nitrogen gas at 27^ ^ ^ C is...

A thermally insulated vessel with nitrogen gas at 27∘C27^{{}^{{}^{\circ}}} \mathrm{C} is moving with a velocity of 100 ms−1100 \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 KE\mathrm{KE} of the gas is given as heat to gas and R=8.3Jmol−1 K−1R=8.3 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1} )

Options

  1. A1.1
  2. B0.93
  3. C0.5
  4. D2.25

Correct answer

D. 2.25

Step-by-step solution

Let there are nn moles of N2N_2 gas in the cylinder. Assuming all of K,EK, E appears in form of heat, n(12Mv2)=f2nRΔTn\left(\frac{1}{2} M v^2\right)=\frac{f}{2} n R \Delta T Here, M=28 g=28×10−3 kg,f=5M=28 \mathrm{~g}=28 \times 10^{-3} \mathrm{~kg}, f=5 Increment in pressure due to change of temperature is Also, Δp=nRΔTV\Delta p=\frac{n R \Delta T}{V} So, Δpp=(nRΔTV)(nRTV)=nRΔTnRT⇒Δpp=nMv2fnRT=Mv2fRT\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 Δpp×\frac{\Delta p}{p} \times100=28×10−3×100×1005×8.3×300×\frac{28 \times 10^{-3} \times 100 \times 100}{5 \times 8.3 \times 300} \times100=2.25 \%
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