Physics - Heat & Thermo Question with Solution | TestHub

Energy possessed by the ideal gas at $27^{\circ }\mathrm{C}$ is $$E_1=3\left(\frac{3}{2}R\times 300\right)=\frac{2700R}{2}$$ Energy possessed by the ideal gas at $227^{\circ }\mathrm{C}$ is $${\mathrm{E}}_2=2\left(\frac{3\mathrm{R}}{2}\times 500\right)=1500\mathrm{R}$$ If $\mathrm{T}$ be the equilibrium temperature, of the mixture, then its energy will be $${\mathrm{E}}_{\mathrm{m}}=5\left(\frac{3\mathrm{RT}}{2}\right)$$ Since, energy remains conserved, $${\mathrm{E}}_{\mathrm{m}}={\mathrm{E}}_1+{\mathrm{E}}_2$$ or $5\left(\frac{3RT}{2}\right)=\frac{2700R}{2}+1500R$ or $\mathrm{T}=380\mathrm{K}$ or $107^{\circ }\mathrm{C}$