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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGEasy2 minQB

PhysicsEasysingle choice

The temperature of a gas is $- 7 8^{\circ} C$ and the average translational kinetic energy of its molecules is $K$ . The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2 K$ is :

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Answer:

Solution:

$K . E = \frac{nf _{1} RT}{2} T_{i} = - 7 8^{\circ} C \to 273 + [- 7 8^{\circ} C] = 195 K$ K.E $α T$ To double the K.E energy temp also become double $T_{f} = 390 K T_{f} = 11 7^{\circ} C$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: QB

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