Two gases occupy two containers A and B the gas in A, of volume 0.10 m 3, exerts a pressure of 1.40 MPa and that in B of volume 0.15 m 3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

Question

Two gases occupy two containers A and B the gas in A, of volume 0.10 m 3, exerts a pressure of 1.40 MPa and that in B of volume 0.15 m 3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa) (A) 0.70 (B) 0.98 (C) 1.40 (D) 210

Tardigrade Admin · Accepted Answer

As the quantity of gas remains constant μA+μB=μ (PA VA/R T)+(PB VB/R T) =(P(VA+VB)/R T) P=(PA VA+PB VB/VA+VB) =(1.4 × 0.1+0.7 × 0.15/0.1+0.15) P=0.98 MPa

0.70

0.98

1.40

210

As the quantity of gas remains constant μA​+μB​=μ RTPA​VA​​+RTPB​VB​​ =RTP(VA​+VB​)​ P=VA​+VB​PA​VA​+PB​VB​​ =0.1+0.151.4×0.1+0.7×0.15​ P=0.98MPa

As the quantity of gas remains constant $μ_{A} + μ_{B} = μ$
$\frac{P _{A} V _{A}}{RT} + \frac{P _{B} V _{B}}{RT}$
$= \frac{P ( V _{A} + V _{B} )}{RT}$
$P = \frac{P _{A} V _{A} + P _{B} V _{B}}{V _{A} + V _{B}}$
$= \frac{1.4 \times 0.1 + 0.7 \times 0.15}{0.1 + 0.15}$
$P = 0.98 MP a$