3 moles of an ideal gas at a temperature of 27^ ^ C are mixed with...

Energy possessed by the ideal gas at $27^{{}^{\circ}} \mathrm{C}$ is E_{1}=3\left(\frac{3}{2}R $\times$300\right)=$\frac{2700 R}{2}$Energy possessed by the ideal gas at $227^{{}^{\circ}} \mathrm{C}$ is $\mathrm{E}_{2}=$2\left(\frac{3 \mathrm{R}}{2} \times500\right)=1500\mathrm{R}$If$\mathrm{T}$be the equilibrium temperature, of the mixture, then its energy will be$\mathrm{E}_{\mathrm{m}}=$5\left(\frac{3 \mathrm{R} \mathrm{T}}{2}\right)$Since, energy remains conserved, $\mathrm{E}_{\mathrm{m}}=\mathrm{E}_{1}+\mathrm{E}_{2}$or $\quad 5\left(\frac{3 R T}{2}\right)=\frac{2700 R}{2}+1500 R$ or $\quad \mathrm{T}=380 \mathrm{~K} or 107^{{}^{\circ}} \mathrm{C}$