# A monoatomic gas of mass \[4.0\mu \]is kept in an insulated container. Container is moving with a velocity\[30m/s\]. If container is suddenly stopped then change in temperature of the gas (\[R = gas{\rm{ }}constant\]) is\[x/3R\]. Value of \[x\]is ______.

Last updated date: 26th Jan 2023

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Answer

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**Hint:**Here, we will use the concept of kinetic energy of gases. The question suggests that given gas is monoatomic and is kept in an insulated container, thus the change in temperature of the gas will be constant. Using the theory behind the problem we will find the value of \[\Delta T\] with \[\dfrac {x}{3R}\].

**Formula Used:**

\[\dfrac{1}{2}m{v^2} = n{C_v}\Delta T\]

\[n = \dfrac{m}{M}\], \[{\rm{m = mass, M = molar\,mass}}\]

\[{C_v} = \dfrac{{3R}}{2}\] \[{{\rm{C}}_{\rm{v}}}{\rm{ = specific\, heat\, capacity\, at \,constant\, volume}}\]

**Complete answer:**

Here, we are equating kinetic energy of the gas and kinetic energy of the box, because molecules in the box are always moving and the box can be moved and stopped but both of them can be equated on the basis of Kinetic theory of gases.

Given,

Mass of the gas is\[4.0\mu \], Velocity of the container is \[30m/s\]

Change in temperature (\[\Delta T\]) is \[x/3R\].

Now, equating the kinetic energy of the box and the kinetic energy of gas, we get

\[\dfrac{1}{2}m{v^2} = n{C_v}\Delta T\]

\[ \Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{m}{M} \times \dfrac{{3R}}{2} \times \Delta T\]

\[\Delta T = \dfrac{{{v^2} \times M}}{{3R}}\] …

(m in RHS is cancelled with m in LHS)

Now, implementing all the required values from the given data, we get

\[ \Rightarrow \Delta T = \dfrac{{{{(30)}^2} \times 4}}{{3R}}\]

Here, in the above equation, mass of the gas was given \[4.0\mu \] but we needed molar mass that is why we used \[4g/mole\] at the place of \[M\].

Therefore,

\[ \Rightarrow \Delta T = \dfrac{{900 \times 4}}{{3R}}\]

\[ \Rightarrow \Delta T = \dfrac{{3600}}{{3R}}\]

Now, we have given change in temperature in the question according to that we have to compare calculated \[\Delta T\] and given \[\Delta T\], we get

\[ \Rightarrow \dfrac{x}{{3R}} = \dfrac{{3600}}{{3R}}\]

\[\therefore x = 3600\]

**Therefore the value of x is 3600.**

**Note:**While solving this particular question we should remember the value of n should be replaced by \[\dfrac {m}{M}\] and the value ove of specific heat at constant volume for monoatomic mass. This will allow us to solve the question without any difficulty.

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