$n_{1}$ and $n_{2}$ moles of two ideal gases (mol. wt. $M_{1}$ and $M_{2}$ ) respectively at temperature $T_{1}$ K and $T_{2}$ K are mixed. Assuming that no loss of energy, the temperature of mixture becomes -

$n_{1} T_{1} + n_{2} T_{2}$

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$\frac{n_{1} T_{1} + n_{2} T_{2}}{T_{1} + T_{2}}$

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$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

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$\frac{n_{1} \times T_{2}}{n_{1} \times n_{2}}$

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Solution

The correct option is C
$\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Let the final temperature be T , then K $E_{1}$ + K $E_{m i x t u r e}$

$\frac{3}{2}$ $n_{1} R T_{1}$ + $\frac{3}{2}$ $n_{2} R T_{2}$ = $\frac{3}{2}$ ( $n_{1} + n_{2}$ ) RT

T = $\frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

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n1 and n2 moles of two ideal gases (mol. wt. M1 and M2) respectively at temperature T1K and T2K are mixed. Assuming that no loss of energy, the temperature of mixture becomes -

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$n_{1}$ and $n_{2}$ moles of two ideal gases (mol. wt. $M_{1}$ and $M_{2}$ ) respectively at temperature $T_{1}$ K and $T_{2}$ K are mixed. Assuming that no loss of energy, the temperature of mixture becomes -