JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, volume of container, V=4.0×10−3m3
Gas constant, R=8.3J‌mol−1K−1
Number of moles of hydrogen, n1=1
Number of moles of carbon dioxide, n2=2
Temperature of mixture, T = 400 K
Using Dalton’s law to calculate the pressure of mixture as follows
p=p1+p2
=

n1RT

V

+

n2RT

V

=

RT

V

(n1+n2) =

RT

V

(1+2)=

3RT

V

=

3×8.3×400

4.0×10−3

=24.9×105‌Pa
Thus, the pressure of mixture of gases is 24.9×105‌Pa.
Thus, the pressure of mixture of gases is 24.9×105‌Pa.