JEE Main Chemistry Class 11 States of Matter Questions

On mixing the temperature of gas becomes T. Internal energy lost by gas A is the internal energy gained by gas B.
ΔUA=ΔUB

f

2

nARΔTA=

f

2

nBRΔTB (As both gases are diatomic, f is same on both sides)
nAΔTA=nBΔTB

2.8

28

×(T1−T)=

0.2

28

(T−T2)

1

10

(300−T)=

1

140

(T−60)
T=284K
On mixing the temperature becomes 284K and volume becomes 3L and total number of moles becomes (nA+nB)=0.10 moles.
pV=(nA+nB)‌RT
=

0.10×8.3×284

3

=84.18‌atm or 84.18‌bar
This is final pressure of gas.