JEE Main · 2011 · 10 Apr 2011 · Offline · Heat & Thermodynamics

Three perfect gases at absolute temperatures

T_{1}, T_{2}

and

T_{3}

are mixed. The masses of molecules are

m_{1}

m_{2}

and

m_{3}

and the number of molecules are

n_{1}, n_{2}

and

n_{3}

respectively. Assuming no loss of energy, the final temperature of the mixture is :

(A) $\frac{n _{1} T _{1} + n _{2} T _{2} + n _{3} T _{3}}{n _{1} + n _{2} + n _{3}}$ ✓ Correct
(B) $\frac{n _{1} T _{1} + n _{2} T _{2}^{2} + n _{3} T _{3}^{2}}{n _{1} T _{1} + n _{2} T _{2} + n _{3} T _{3}}$
(C) $\frac{n _{1}^{2} T _{1}^{2} + n _{2}^{2} T _{2}^{2} + n _{3}^{2} T _{3}^{2}}{n _{1} T _{1} + n _{2} T _{2} + n _{3} T _{3}}$
(D) $\frac{( T _{1} + T _{2} + T _{3} )}{3}$

Answer: (A)

Solution

With no energy lost in mixing, the total internal energy before equals that after. For ideal gases the internal energy of each sample is proportional to its number of molecules times its temperature, so

n_{1} k T_{1} + n_{2} k T_{2} + n_{3} k T_{3} = (n_{1} + n_{2} + n_{3}) k T

. Solving for the common final temperature,

T = \frac{n _{1} T _{1} + n _{2} T _{2} + n _{3} T _{3}}{n _{1} + n _{2} + n _{3}}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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JEE Main · 2011 · 10 Apr 2011 · Offline · Heat & Thermodynamics

Solution

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