Ncert Exercises & Solutions

An insulated container containing monoatomic gas of molar mass m is moving with a velocity, $v_{o}$ $v_{o}$ . If the container is suddenly stopped, find the change in temperature.

Hint: The kinetic interpretation of temperature, the absolute temperature of a given sample of a gas is proportional to the total translational kinetic energy of its molecules.

Step 1: Find the change in translational KE of the gas.
Any change in the absolute temperature of a gas will contribute to the corresponding change in translational kinetic energy and vice-versa.
Assuming, n = number of moles
Given, m = molar mass of the gas
When the container stops, its total KE is transferred to gas molecules in the form of
translational KE, thereby increasing the absolute temperature.
If

∆

$∆$ T = change in absolute temperature

Then, KE of molecules due to velocity

v_{0}, K E = \frac{1}{2} (m n) v_{0}^{2}

$v_{0}, K E = \frac{1}{2} (m n) v_{0}^{2}$ ...(i)

Increase in translational KE

= n \frac{3}{2} R (∆ T)

$= n \frac{3}{2} R (∆ T)$ ...(ii)

Step 2: Find the change in temperature.

According to kinetic theory, Eqs. (i) and (ii) are équal.

\Rightarrow \frac{1}{2} (m n) v_{0}^{2} = \frac{3}{2} n R (∆ T)

$\Rightarrow \frac{1}{2} (m n) v_{0}^{2} = \frac{3}{2} n R (∆ T)$

(m n) v_{0}^{2} = 3 n R (∆ T)

$(m n) v_{0}^{2} = 3 n R (∆ T)$

\Rightarrow ∆ T = \frac{(m n) v_{0}^{2}}{3 n R}

$\Rightarrow ∆ T = \frac{(m n) v_{0}^{2}}{3 n R}$

\Rightarrow ∆ T = \frac{m v_{0}^{2}}{3 R}

$\Rightarrow ∆ T = \frac{m v_{0}^{2}}{3 R}$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions