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PhysicsKTGBasic of KTG/Gas lawsMedium2 minPYQ_2025

PhysicsMediumsingle choice

The kinetic energy of translation of the molecules in 50 g of $CO_{2}$ gas at $1 7^{\circ} C$ is

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Answer:

Solution:

Kinetic energy of translation $= \frac{3}{2} n RT$
$n = \frac{50 g}{44 g} = \frac{25}{22} mol T = 1 7^{\circ} C = 290 K$
$\Rightarrow$ Kinetic energy of translation
$= \frac{3}{2} (\frac{25}{22}) (8.3) (290) J = 4102.8 J$

Stream:JEESubject:PhysicsTopic:KTGSubtopic:Basic of KTG/Gas laws

⏱ 2mℹ️ Source: PYQ_2025

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