Question

An insulated container containing monatomic gas of molar mass moving with a velocity ${v_ \circ }$.If the container is suddenly stopped. The change in temperature is?
A.$\dfrac{{m{v_ \circ }^2}}{{2R}}$
B.$\dfrac{{m{v_ \circ }^2}}{{3R}}$
C.$\dfrac{R}{{m{v_ \circ }^2}}$
D.$\dfrac{{3m{v_ \circ }^2}}{{2R}}$

Answer 1

Hint: Absolute temperature of a given sample of a gas is directly proportional to its total translational kinetic energy of its molecules according to kinetic interpretation of temperature.

Complete answer:
As we know that absolute temperature of a given sample of a gas is directly proportional to its total translational kinetic energy of its molecules. Hence any change in temperature of a gas will contribute to change in translational kinetic energy and vice-versa.
We are given the molar mass of monatomic gas as $m$ .
If the number of moles of gas is $n$ and change in absolute temperature is $\vartriangle t$ .
Then the mass of molecules is equal to $mn$ .
According to the situation we stop the container so the kinetic energy of the container will transfer in the form of translational kinetic energy to its molecules, thereby increment will occur in temperature.
Now we calculate the KE of the gas container.
$K.E = \dfrac{1}{2}\left( {Mass} \right){\left( {Velocity} \right)^2}$
KE of molecules due to initial velocity, $\dfrac{1}{2}\left( {mn} \right){\left( {{V_ \circ }} \right)^2}$ and it will be named as equation $1$
Now increase in translational KE of monoatomic gas, $\dfrac{3}{2}nR\vartriangle T$ and it will be named as equation $2$
Here $R$ is Universal gas constant, $\Delta T$ is change in absolute temperature, $n$ is number of moles of gas.
According to kinetic theory $Equation(1) = Equation(2)$ .
$ \Rightarrow \dfrac{1}{2}(mn){({v_0})^2} = \dfrac{3}{2}nR(\Delta T)$
After solving it, we get
$ \Rightarrow mn{({v_0})^2} = 3nR(\Delta T)$
Further solving more,
$ \Rightarrow m{v_0}^2 = 3R(\Delta T)$
We get
$ \Rightarrow \Delta T = \dfrac{{m{v_0}^2}}{{3R}}$
Hence the change in temperature of gas is $\dfrac{{m{v_0}^2}}{{3R}}$ .

So, option B is correct.

Note:
The Total energy of the system is conserved; neither we can produce energy nor we can destroy the energy; it only can change its form only. Here in this situation the Kinetic Energy of the system converts into heat that is also a form of Energy.