JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2021
- Hard
- MCQ
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
Pick an option to check your answer.
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.
\(\Delta U =0\)
\(\Rightarrow \frac{ F _{1}}{2} n _{1} R \Delta T +\frac{ F _{2}}{2} n _{2} R \Delta T =0\)
\(\Rightarrow \frac{ F _{1}}{2} n _{1} R \left( T _{1}- T \right)+\frac{ F _{2}}{2} n _{2} R \left( T _{2}- T \right)=0\)
\(\Rightarrow T =\frac{ F _{1} n _{1} RT _{1}+ F _{2} n _{2} RT _{2}}{ F _{1} n _{1} R + F _{2} n _{2} R } \Rightarrow \frac{ F _{1} n _{1} T _{1}+ F _{2} n _{2} T _{2}}{ F _{1} n _{1}+ F _{2} n _{2}}\)
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