Question

# Two gases occupy two containers A and B. The gas in A, of volume 0.10 m3, exerts a pressure of 1.40 MPa and that in B, of volume 0.15 m3, exerts a pressure of 0.7 MPa. The two containers are joined by a tube of negligible volume and the gases are allowed to intermingle. Then, if the temperature remains constant, the final pressure in the containers will be ( in MPa):

A
0.70
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.98
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1.40
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 0.98Given, Volume of gas in container A (VA)=0.10 m3 Volume of gas in container B (VB)=0.15 m3 Pressure exerted by gas in container A (PA)=1.40 MPa Pressure exerted by gas in container B (PB)=0.7 MPa Let P and V be the pressure and volume of the gas after the two containers are joined together. As the quantity of gas remains constant, n=nA+nB Since the temperature is kept constant, by using ideal gas equation, we can write that, P(VA+VB)RT=PAVART+PBVBRT (∵V=VA+VB) ⇒P=PAVA+PBVBVA+VB=1.40×0.1+0.7×0.150.1+0.15 ⇒P=0.98 MPa Thus, option (b) is the correct answer.

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Explore more
Join BYJU'S Learning Program
Select...