### JEE Main Solved Paper 2017 Question 2

##### Question: The temperature of an open room of volume $ 30m^{3} $ increases from $ 17{{,}^{o}}C $ to $ 27{{,}^{o}}C $ due to sunshine. The atmospheric pressure in the room remains $ 1\times 10^{5},Pa. $ If $ n _{i} $ and $ n _{f} $ are the number of molecules in the room before and after heating, then $ n _{f}-n _{i} $ will be:- [JEE Main Solved Paper-2017]

#### Options:

A) $ 2.5\times 10^{25} $

B) $ -2.5\times 10^{25} $

C) $ -1.61\times 10^{23} $

D) $ 1.38\times 10^{23} $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

- Using ideal gas equation PV = NRT (N is number of moles) $ P _0V _0=N _{i}R\times 290 $ ?.. $ [T _{i}=237+17=290K] $ After heating $ P _0V _0=N _{f}R\times 300 $ ?.. from equation [a] and [b] $ N _{f}-N _{i}=\frac{P _0V _0}{R}[ \frac{10}{290\times 300} ] $ Hence $ n _{f}-n _{i} $ is $ =-\frac{P _0V _0}{R}\times [ \frac{10}{290\times 300} ]\times 6.023\times 10^{23} $ Putting $ P _0=10^{5}P _{A} $ and $ V _0=30m^{3} $ Number of molecules $ n _{f}-n _{i}=-2.5\times 10^{25} $