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

Energy possessed by the ideal gas which is at 27oC is E1=3×internal energy in the form of KEE1=3×32RT=332×R(273+27)  = 2700R2Energy possessed by the ideal gas which is at 227oC isE2=23R2×(273+227)=1500RIf T be the equilibrium temperature of the mixture, then its energy will be  Em=53RT2Since, energy remains conserved, Em=E1+E25  3RT2=2700  R2+1500R⇒T=380 K or T=107°C