Question

An insulated box containing monatomic ideal gas of molar mass M is moving with a uniform speed v. The box suddenly stops and consequently the gas acquires a new temperature. Calculate the change in the temperature of the gas. Neglect heat absorbed by the box:
A. $\Delta T = 2\dfrac{{M{v^2}}}{{3R}}$
B. $\Delta T = \dfrac{{M{v^2}}}{{3R}}$
C. $\Delta T = 3\dfrac{{M{v^2}}}{{3R}}$
D. $\Delta T = 4\dfrac{{M{v^2}}}{{3R}}$

Answer 1

Hint:-The box is insulated which means that no heat will be absorbed from the outside of the box. The change in the kinetic energy of the box will be the reason for increase in the internal energy and also the change in the kinetic energy.
Formula used: The formula of the kinetic energy of a body of mass ‘m’ and velocity ‘v’ is given by,
$K.E = \dfrac{1}{2}m \cdot {v^2}$………eq. (1)
The formula of increase in internal energy of the monatomic gas is given by,
$\Delta E = \dfrac{3}{2}nR\Delta T$………eq. (2)
Where n is the number of moles R is the universal gas constant and $\Delta T$ is the change in temperature.

Complete step-by-step solution:It is given that an insulated box containing monatomic ideal gas of molar mass M is moving with a uniform speed v the box suddenly stops and consequently the gas acquires a new temperature and we need to find the change in the temperature of the gas.
The change in the internal energy is equal to the increase in the internal energy of the monatomic gas.
As the final velocity of the box is zero and therefore the final kinetic energy of the box will be zero.
Since the change in the kinetic will be equal to the increase in the internal energy of the monatomic gas in the box.
$ \Rightarrow {\text{K}}{\text{.E}} = {\text{Increase in internal energy }}$
Using equation (1) and equation (2) we get,
$ \Rightarrow \dfrac{1}{2}m \cdot {v^2} = \dfrac{3}{2}nR\Delta T$………eq. (3)
As mole is defined as the ratio of mass of the gas to the molecular mass of the gas so we get,
$n = \dfrac{m}{M}$.
Replace the value of n in the equation (3)
$ \Rightarrow \dfrac{1}{2}m \cdot {v^2} = \dfrac{3}{2}nR\Delta T$
$ \Rightarrow \dfrac{1}{2}m \cdot {v^2} = \dfrac{3}{2}\left( {\dfrac{m}{M}} \right)R\Delta T$
$ \Rightarrow {v^2} = \left( {\dfrac{3}{M}} \right)R\Delta T$
$ \Rightarrow \Delta T = \dfrac{{M \cdot {v^2}}}{{3 \cdot R}}$
The change in the temperature of the gas is equal to$\Delta T = \dfrac{{M \cdot {v^2}}}{{3 \cdot R}}$.

The correct answer for this problem is option B.

Note:- The final velocity of the box is zero and therefore the value of kinetic energy will be also zero. The change in internal energy of a gas is caused by the striking of the molecules of gas with each other and also with the walls of the container.