JEE Main 12 Jan 2019 Evening Question 20
Question: An open vessel $ 27{}^\circ C $ at is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 750 K
B) $ 750{}^\circ C $
C) $ 500{}^\circ C $
D) 500 K
Show Answer
Answer:
Correct Answer: D
Solution:
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From ideal gas equation, $ \frac{P _1V _1}{n _1RT _1}=\frac{P _2V _2}{n _2RT _2} $ $ \because $ $ P _1V _1=P _2V _2; $
$ \therefore n _1T _1=n _2T _2 $ $ n _1\times 300K=\frac{3}{5}n _1\times T _2 $ $ T _2=500K $
