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JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases

Two ideal polyatomic gases at temperatures \(T _{1}\) and \(T _{2}\) are mixed so that there is no loss of energy. If \(F _{1}\) and \(F _{2}, m _{1}\) and \(m _{2}, n _{1}\) and \(n _{2}\) be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is

  1. A \(\frac{ n _{1} T _{1}+ n _{2} T _{2}}{ n _{1}+ n _{2}}\)
  2. B \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ n _{1} F _{1}+ n _{2} F _{2}}\)
  3. C \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ F _{1}+ F _{2}}\)
  4. D \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ n _{1}+ n _{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ n _{1} F _{1}+ n _{2} F _{2}}\)

Step-by-step Solution

Detailed explanation

Let the final temperature of the mixture be \(T\). Since, there is no loss in energy. \(\Delta U =0\) \(\Rightarrow \frac{ F _{1}}{2} n _{1} R \Delta T +\frac{ F _{2}}{2} n _{2} R \Delta T =0\)…
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