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Question: Two gases occupy two containers A and B the gas in A, of volume \(0.10m^{3}\), exerts a pressure of ...

Two gases occupy two containers A and B the gas in A, of volume 0.10m30.10m^{3}, exerts a pressure of 1.40 MPa and that in B of volume 0.15m30.15m^{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 it the temperature remains constant, the final pressure in the container will be (in MPa)

A

(a) 0.70

B

0.98

C

1.40

D

2.10

Explanation

Solution

(a)

Dalton’s law =RTV[m1M1+m2M2+m3M3]=8.31×3003×10−3[632+828+544]=498×103=~500×103= \frac{RT}{V}\left\lbrack \frac{m_{1}}{M_{1}} + \frac{m_{2}}{M_{2}} + \frac{m_{3}}{M_{3}} \right\rbrack = \frac{8.31 \times 300}{3 \times 10^{- 3}}\left\lbrack \frac{6}{32} + \frac{8}{28} + \frac{5}{44} \right\rbrack = 498 \times 10^{3}\widetilde{=}500 \times 10^{3}.=~5×105N⥂⥂/⥂m2\widetilde{=}5 \times 10^{5}N ⥂ ⥂ / ⥂ m^{2}