Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGEasy2 minPYQ_2024

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The total kinetic energy of $1$ mole of oxygen at $27 ° C$ is :
[Use universal gas constant $(R) = 8.31 J {mol}^{- 1} K^{- 1}$ ]

Answer:

Solution:

The formula to calculate the kinetic energy can be written as

$K = \frac{f}{2} n R T . . . (1)$

Since oxygen is a diatomic gas, the number of degrees of freedom is $5$ .

Thus, from equation (1), it follows that

$\begin{array}{rcl} K & = & \frac{5}{2} \times 1 \times 8.31 \times 300 J \\ = & 6232.5 J \end{array}$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: PYQ_2024

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