3 moles of an ideal monoatomic gas at a temperature of 27^oC are mixed with 2  moles of an ideal monoatomic gas at a temperature 227^oC. Determine the  equilibrium temperature of the mixture, assuming no loss of energy.

Energy possessed by the ideal gas which is at 27^oC is 

$E_1=\text{3}\times \text{internal energy in the form of KE}$

$E_1=3\times \left(\frac{3}{2}RT\right)=3\left(\frac{3}{2}\times R(273+27)\right)=\frac{2700R}{2}$

Energy possessed by the ideal gas which is at 227^oC is

$E_2=2\left(\frac{3R}{2}\times (273+227)\right)=1500R$

If T be the equilibrium temperature of the mixture, then its energy will be  $E_m=5\left(\frac{3RT}{2}\right)$

Since, energy remains conserved, $E_m=E_1+E_2$

$5\left(\frac{3RT}{2}\right)=\frac{2700R}{2}+1500R$

$\Rightarrow \text{T}=\text{38}0\text{ K or T}=\text{1}0\text{7}°\text{C}$