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BITSAT 2022 Paper

Section: Physics

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Question : 18 of 130

Marks: +1, -0

A monoatomic gas of molar mass

m

is kept in a insulated container. Container is moving with velocity

v

. If the container is suddenly stopped, then the change in the temperature of the gas is

$\frac{m v^2}{4 R}$
$\frac{m v^2}{2 R}$
$\frac{m v^2}{R}$
$\frac{m v^2}{3 R}$

Solution: 👈: Video Solution

Let the insulated container has n moles of the monoatomic gas. The loss of kinetic energy of the gas,

\Delta E_K = \frac{1}{2} (m v) v^2

If the change in temperature of the gas is

\Delta T

, then heat gained by the gas is,

\Delta Q = \frac{3}{2} nR \Delta T

Now, according to question,

\Delta Q = \Delta E_K

\frac{3}{2} nR \Delta T = \frac{1}{2} m n v^2

\Rightarrow \Delta T = \frac{m v^2}{3 R}

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