Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGEasy2 minPYQ_2020

PhysicsEasysingle choice

The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at $127 ° C$ is

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Answer:

Solution:

Root mean square velocity

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$

where $R$ is gas constant, $T$ the temperature and $M$ molecular weight.

Given, ${M_{N}}_{2} = 28, {M_{O}}_{2} = 32, {T_{O}}_{2} = 127 ℃ = 127 + 273 = 400$ K

$∴ \frac{{v_{O}}_{2}}{{v_{N}}_{2}} = \sqrt{\frac{{T_{O}}_{2}}{{M_{O}}_{2}} \times \frac{{M_{N}}_{2}}{{T_{N}}_{2}}} = \sqrt{\frac{400}{32} \times \frac{28}{{T_{N}}_{2}}} = 1$

$\Rightarrow {T_{N}}_{2} = 350 K = 77 ℃$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: PYQ_2020

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