Question

A sample of gas in a box is at pressure \[{P_0}\] and temperature\[{T_0}\]. If number of molecules is doubled and total kinetic energy of the gas kept constant, then final temperature and pressure will be
\[{T_0}.\,{P_0}\]
\[{T_0}.\,2{P_0}\]
\[\dfrac{{{T_0}}}{2}.\,2{P_0}\]
\[\dfrac{{{T_0}}}{2}.\,{P_0}\]

Answer 1

Hint: Use the formula for kinetic energy of the gas relating number of moles and temperature to find the relation between final and initial temperature. In ideal gas equation, the product of pressure and volume remains constant.

Formula Used: \[K.E = \dfrac{f}{2}nR{T_0}\]
Here, f is the degree of freedom, n is the number of moles of the gas, and R is the gas constant.

Complete step by step answer:We aren't given whether the gas is monatomic or diatomic. Therefore, the kinetic energy of the gas at temperature \[{T_0}\] is,
\[K.E = \dfrac{f}{2}nR{T_0}\]

Here, f is the degree of freedom, n is the number of moles of the gas, and R is the gas constant.

The kinetic energy of the gas when the number of molecules is doubled is,
\[K.E = \dfrac{f}{2}\left( {2n} \right)R{T_1}\]

Here, \[{T_1}\] is temperature of the gas when the number of molecules is doubled.

We have given that the kinetic energy of the gas is constant, therefore, we can write,
\[\dfrac{f}{2}\left( {2n} \right)R{T_1} = \dfrac{f}{2}nR{T_0}\]
\[ \Rightarrow {T_1} = \dfrac{{{T_0}}}{2}\]

Therefore, the temperature of the gas will be half of the initial value.

The volume of the box is constant. Therefore, the volume of the gas will not increase in both the cases.

Now, according to the ideal gas equation, \[PV = nRT\], the product \[PV\] always remains constant. Since the volume does not change, the pressure also must not change.

Therefore, the final pressure will be \[{P_0}\].

So, the correct answer is option (D).

Note:Students should note that when the number of molecules is doubled, the number of moles also gets doubled. We know that one mole is equal to Avagadro’s number. When the number of moles of the gas increases, the temperature of the gas decreases, keeping the product \[PV\] constant.