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Question: Two moles of a gas A at 27°C mixed with a 3 moles of gas at 37°C. If both are monoatomic ideal gases...

Two moles of a gas A at 27°C mixed with a 3 moles of gas at 37°C. If both are monoatomic ideal gases, what will be the temperature of the mixture ?

A

66°C

B

11°C

C

22°C

D

33°C

Answer

33°C

Explanation

Solution

Since there is no loss of energy in the process, therefore, sum of kinetic energy of gases A and B = Kinetic energy of mixture.

Temperature of the mixture, T=μ1T1+μ2T2μ1+μ2T = \frac{\mu_{1}T_{1} + \mu_{2}T_{2}}{\mu_{1} + \mu_{2}}

=2(27+273)+3(37+273)2+3= \frac{2(27 + 273) + 3(37 + 273)}{2 + 3}

=600+9305=15305= \frac{600 + 930}{5} = \frac{1530}{5}

∴=360K=360−273=33∘C\therefore = 360K = 360 - 273 = 33{^\circ}C