# An open vessel containing air is heated from 300K to 400K. The fraction of air, which goes out with respect to originally present is:

PV = nRT , T1 = 300k , T2 = 400k

Since, it is an open vessel, pressure is constant and there is no change in the volume of vessel.

$\therefore$   P1 = P2  and V1 = V2

$\frac{n_{1}}{n_{2}} = \frac{T_{2}}{T_{1}} = \frac{400}{300} = \frac{4}{3}$

$n_{2} = \frac{3}{4}n_{1}$

Therefore, fraction of air that goes out $= 1 - \frac{3}{4} = \frac{1}{4}$

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-