Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGEasy2 minPYQ_2020

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The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the $r m s$ speed of the gas molecule is $V$ at $100 K$ , then at $400 K$ it becomes

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Answer:

Solution:

$∵ r m s$ speed of molecule $V_{r m s} \propto \sqrt{T}$

for $T = 100 K$

${(V_{r m s})}_{1} \propto \sqrt{100}$

For $T = 400 K$

${(V_{r m s})}_{2} \propto \sqrt{400}$

therefore, $\frac{{(V_{r m s})}_{1}}{{(V_{r m s})}_{2}} = \frac{\sqrt{100}}{\sqrt{400}}$

or $\frac{V}{{(V_{r m s})}_{2}} = \frac{1}{2}$

or ${(V_{r m s})}_{2} = 2 V$

Subject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: PYQ_2020

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