Question
Two moles of gas A at 27°C are mixed with 3 moles of gas B at 37°C. If both are monoatomic ideal gases, what will be the temperature of the mixture?
✓
Answer
As there is no loss of energy in the process, therefore, sum of KE of gases A and B = KE of mixture,$\mu_1\Big(\frac{3}{2}\text{RT}_1\Big)+\mu_2\Big(\frac{3}{2}\text{RT}_2\Big)$
$=(\mu_1+\mu_2)\frac{3}{2}\text{RT}$
where T is temperature of the mixture.$\therefore\text{T}=\frac{\mu_1\text{T}_1+\mu_2\text{T}_2}{\mu_1+\mu_2}$
$=\frac{2(27+273)+3(37+273)}{2+3}$
$=\frac{600+930}{5}=\frac{1530}{5}$
$=306-273=3^{\circ}\text{C}$
$=(\mu_1+\mu_2)\frac{3}{2}\text{RT}$
where T is temperature of the mixture.$\therefore\text{T}=\frac{\mu_1\text{T}_1+\mu_2\text{T}_2}{\mu_1+\mu_2}$
$=\frac{2(27+273)+3(37+273)}{2+3}$
$=\frac{600+930}{5}=\frac{1530}{5}$
$=306-273=3^{\circ}\text{C}$
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