Consider a mass  $m=15gm$ of nitrogen enclosed in a vessel at temperature  $T=300K$.   The amount of heat required to double the root mean square velocity of its molecules  nearest integer N in ${10}^N$  Joules is

All the diatomic gases $(H_2,O_2,N_2,etc.)$  exhibit 5 degrees of freedom between 
100 – 1000 K, 3 of translation and 2 of rotation. Hence, internal energy of $n$  moles of $N_2$  at T = 300 K would be $U=\frac{5}{2}nRT$
Translational kinetic energy of each molecule is
 $\frac{1}{2}{m}_a{v}_{rms}^2=\frac{3}{2}kT$ 
where $m_a$   is the mass of the molecule.
If rms speed increases $\alpha$  times, ${v^{\text{'}}}_{rms}=\alpha v_{rms}$ , then temperature has to be raised to $T^{\text{'}}$  such that
$\frac{1}{2}{m}_a{v^{\text{'}}}_{rms}^2=\frac{3}{2}k{T}^{\text{'}}$  
or  $T^{\text{'}}={\alpha }^{2}T$
The internal energy therefore increases to
$U^{\text{'}}=\frac{5}{2}nR{T}^{\text{'}}$  
Change in internal energy,
 $$\begin{gathered} \Delta U=U^{\text{'}}-U=\frac{5}{2}nR(T^{\text{'}}-T) \\ =\frac{5}{2}nRT({\alpha }^2-1) \\ =\frac{5}{2}\frac{m}{M}RT({\alpha }^2-1) \end{gathered}$$ 
where the number of moles $n=\frac{m}{M}$ ,  $m$ is the mass of gas and M is its molecular weight.
On substituting numerical values, $m=15g$ , $T=300K$ ,$\alpha =2$   and $M=28g/mol$ for  $N_2$, we get $U=\frac{5}{2}\times \frac{15}{28}\times 8.31\times 300\times (4-1)J$$={10}^{4}J$