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JEE Main 2011
Exam
JEE Main
Year
2011
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2011 · Physics · Kinetic Theory of Gases

Three perfect gases at absolute temperatures T_1, T_2 and T_3 are...

Three perfect gases at absolute temperatures T1,T2T_1, T_2 and T3T_3 are mixed. The masses of molecules are m1m_1, m2m_2 and m3m_3 and the number of molecules are n1,n2n_1, n_2 and n3n_3 respectively. Assuming no loss of energy, the final temperature of the mixture is :

Options

  1. An1T1+n2T2+n3T3n1+n2+n3\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}
  2. Bn1T1+n2T22+n3T32n1T1+n2T2+n3T3\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}
  3. Cn12T12+n22T22+n32T32n1T1+n2T2+n3T3\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}
  4. D\frac{\left(T_1+T_2+T_3\right)}{3}

Correct answer

A. n1T1+n2T2+n3T3n1+n2+n3\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}

Step-by-step solution

Data ⇒n,k,t1+n2kT2+n3kT3=(n1+n2+n3)kT\Rightarrow n, k, t_1+n_2 k T_2+n_3 k T_3=\left(n_1+n_2+n_3\right) k T ∴T=n1T1+n2T2+n3T3n1+n2+n3\therefore T=\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}
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