Three perfect gases at absolute temperatures T1,T2 and T3 are mixed. The masses of molecules are m1,m2 and m3 and the number of molecules are n1,n2 and n3 respectively. Assuming no loss of energy, the final temperature of the mixture is

As Degrees of freedom is not stated, we will assume Cv is same for all gases Ui=Uf⇒n1CvT1+n2CvT2+n3CvT3=n1+n2+n3CvT⇒T=n1T1+n2T2+n3T3n1+n2+n3

Three perfect gases at absolute temperatures T1,T2 and T3 are mixed. The masses of molecules are m1,m2 and m3 and the number of molecules are n1,n2 and n3 respectively. Assuming no loss of energy, the final temperature of the mixture is

As Degrees of freedom is not stated, we will assume Cv is same for all gases Ui=Uf⇒n1CvT1+n2CvT2+n3CvT3=n1+n2+n3CvT⇒T=n1T1+n2T2+n3T3n1+n2+n3

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Three perfect gases at absolute temperatures T₁,T₂ and T₃ are mixed. The masses of molecules are m₁,m₂ and m₃ and the number of molecules are n₁,n₂ and n₃ respectively. Assuming no loss of energy, the final temperature of the mixture is

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$\frac{n_{1} T_{1}^{2} + n_{2} T_{2}^{2} + n_{3} T_{3}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$

$\frac{n_{1}^{2} T_{1}^{2} + n_{2}^{2} T_{2}^{2} + n_{3}^{2} T_{3}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{(T_{1} + T_{2} + T_{3})}{3}$

answer is A.

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Detailed Solution

As Degrees of freedom is not stated, we will assume C_v is same for all gases $U_{i} = U_{f}$
$\begin{array}{l} \Rightarrow n_{1} C_{v} T_{1} + n_{2} C_{v} T_{2} + n_{3} C_{v} T_{3} = (n_{1} + n_{2} + n_{3}) C_{v} T \\ \Rightarrow T = \frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}} \end{array}$