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
- A \(\frac{ n _{1} T _{1}+ n _{2} T _{2}}{ n _{1}+ n _{2}}\)
- B \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ n _{1} F _{1}+ n _{2} F _{2}}\)
- C \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ F _{1}+ F _{2}}\)
- D \(\frac{ n _{1} F _{1} T _{1}+ n _{2} F _{2} T _{2}}{ n _{1}+ n _{2}}\)
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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