Kinetic Theory Of Gases Question 146
Question: Three perfect gases at absolute temperatures $ T _{1},,T _{2} $ and $ T _{3} $ are mixed. The masses of molecules are $ m _{1},,m _{2} $ and $ m _{3} $ and the number of molecules are $ n _{1},,n _{2} $ and $ n _{3}, $ respectively. Assuming no loss of energy, the final temperature of the mixture is
Options:
A) $ \frac{(T _{1}+T _{2}+T _{3})}{3} $
B) $ \frac{n _{1}T _{1}+n _{2}T _{2}+n _{3}T _{3}}{n _{1}+n _{2}+n _{3}} $
C) $ \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}} $
D) $ \frac{n _{1}^{2}T _{1}^{2}+n _{3}^{2}T _{2}^{2}+n _{3}^{2}T _{3}^{2}}{n _{1}T _{1}+n _{2}T _{2}+n _{3}T _{3}} $
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Answer:
Correct Answer: B
Solution:
[b] Let $ T _{3}>T _{2}>T _{1} $ and final temperature is T such that $ T _{3}>T>T _{2}>T _{1} $ . Now heat gained by first tow gases is equal to heat lost by third gas $ n _{1}C(T-T _{1})+n _{2}C(T-T _{2})=n _{3}C(T _{3}-T) $
$ \Rightarrow $ $ T=\frac{n _{1}T _{1}+n _{2}T _{2}+n _{3}T _{3}}{n _{1}+n _{2}+n _{3}} $
BETA
