## In This post you will get answer of Q.

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

Solution:

## Energy possessed by the ideal gas at 27°C is

$E_{1} = 3 left(frac{3}{2} R times300right) = left(frac{2700 R}{2}right)$

Energy possessed by the ideal gas at 227°C is

$E_{2} = 2left(frac{3R}{2} times500right) = 1500;R$

If T be the equilibrium temperature, of the

mixture, then its energy will be

$E_{2} = 5left(frac{3RT}{2}right)$

Since, energy remains conserved,

$E_m:=E_1:+:E_2:$

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

$:or:::::T:=:380:K:or:107^{circ }:C:$