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

Step-by-step solution

Energy possessed by the ideal gas at 27^ C is E₁=3 ( 3 2 R 300 )= 2700 R 2 Energy possessed by the ideal gas at 227^ C is E ₂=2 ( 3 R 2 500 )=1500 R If T be the equilibrium temperature, of the mixture, then its energy will be E _ m =5 ( 3 R T 2 ) Since, energy remains conserved, E _ m = E ₁+ E ₂ or 5 ( 3 R T 2 )= 2700 R 2 +1500 R or T =380 ~K or 107^ C