Kinetic Theory

JEE Main level, single correct, medium

147 of 168

Question 147 of 168

Two ideal polyatomic gases at temperatures T1{T}_{1} and T2{T}_{2} are mixed so that there is no loss of energy. If F1{F}_{1} and F2,m1{F}_{2},{m}_{1} and m2,n1{m}_{2},{n}_{1} and n2{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:

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Answer: (B) n1F1T1+n2F2T2n1F1+n2F2\dfrac{{n}_{1}{F}_{1}{T}_{1}+{n}_{2}{F}_{2}{T}_{2}}{{n}_{1}{F}_{1}+{n}_{2}{F}_{2}}

Let the final temperature of the mixture be T.T.
Since, there is no loss in energy.
ΔU=0\Delta U=0
⇒F12n1RΔT+F22n2RΔT=0\Rightarrow \dfrac{{F}_{1}}{2}{n}_{1}R\Delta T+\dfrac{{F}_{2}}{2}{n}_{2}R\Delta T=0
⇒F12n1R(T1−T)+F22n2R(T2−T)=0\Rightarrow \dfrac{{F}_{1}}{2}{n}_{1}R\left({T}_{1}-T\right)+\dfrac{{F}_{2}}{2}{n}_{2}R\left({T}_{2}-T\right)=0
⇒T=F1n1RT1+F2n2RT2F1n1R+F2n2R=F1n1T1+F2n2T2F1n1+F2n2\Rightarrow T=\dfrac{{F}_{1}{n}_{1}R{T}_{1}+{F}_{2}{n}_{2}R{T}_{2}}{{F}_{1}{n}_{1}R+{F}_{2}{n}_{2}R}=\dfrac{{F}_{1}{n}_{1}{T}_{1}+{F}_{2}{n}_{2}{T}_{2}}{{F}_{1}{n}_{1}+{F}_{2}{n}_{2}}

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