2 text{ mole} of N_2 gas at 27 , ^oC is mixed | vedclass

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(B) The final temperature of a mixture of non-reacting gases is given by the formula: $T_{	ext{mix}} = \frac{n_1 C_{v1} T_1 + n_2 C_{v2} T_2}{n_1 C_{

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$277 \, K$

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(B) The final temperature of a mixture of non-reacting gases is given by the formula: $T_{	ext{mix}} = \frac{n_1 C_{v1} T_1 + n_2 C_{v2} T_2}{n_1 C_{v1} + n_2 C_{v2}}$.For $N_2$ (diatomic gas),$n_1 = 2 	ext{ mole}$ and $C_{v1} = \frac{5}{2}R$. The temperature $T_1 = 27 + 273 = 300 \, K$.For $He$ (monatomic gas),$n_2 = 1 	ext{ mole}$ and $C_{v2} = \frac{3}{2}R$. The temperature $T_2 = -73 + 273 = 200 \, K$.Substituting the values:$T_{	ext{mix}} = \frac{2 	imes (\frac{5}{2}R) 	imes 300 + 1 	imes (\frac{3}{2}R) 	imes 200}{2 	imes (\frac{5}{2}R) + 1 	imes (\frac{3}{2}R)}$$T_{	ext{mix}} = \frac{1500R + 300R}{5R + 1.5R} = \frac{1800R}{6.5R} = \frac{1800}{6.5} \approx 276.92 \, K \approx 277 \, K$.

$2 \text{ mole}$ of $N_2$ gas at $27 \, ^oC$ is mixed with $1 \text{ mole}$ of $He$ gas at $-73 \, ^oC$. Find the final temperature $T_{\text{mix}}$ of the mixture.

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