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(B) The total kinetic energy of an ideal gas is given by $U = \frac{3}{2} N k_B T$,where $N$ is the number of molecules,$k_B$ is the Boltzmann constan

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$\frac{T}{2}$

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(B) The total kinetic energy of an ideal gas is given by $U = \frac{3}{2} N k_B T$,where $N$ is the number of molecules,$k_B$ is the Boltzmann constant,and $T$ is the absolute temperature.Initially,the total kinetic energy is $U_i = \frac{3}{2} N k_B T$.Finally,the number of molecules becomes $N' = 2N$ and the total kinetic energy remains the same,so $U_f = U_i$.Let the new temperature be $T'$. Then $U_f = \frac{3}{2} (2N) k_B T'$.Equating the energies: $\frac{3}{2} N k_B T = \frac{3}{2} (2N) k_B T'$.Simplifying the equation: $T = 2T'$.Therefore,the new temperature is $T' = \frac{T}{2}$.

$A$ container has $N$ molecules at absolute temperature $T$. If the number of molecules is doubled but the total kinetic energy in the box remains the same as before,the absolute temperature of the gas is ...........

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