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The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 minutes when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 79 in the molar ratio of 1 : 1 at a total pressure of 4000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 minutes.

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra
NOTE THIS STEP : Since the evacuated bulb contains a mixture of oxygen and another gas in the molar ratio of 1 : 1 at a total pressure of 4000 mm, the partial pressure of each gas is 2000 mm.
The drop in the pressure of oxygen after 74 minutes
= 500/47 = 787.2 mm of Hg
∴ After 74 minutes, the pressure of oxygen
= 2000 – 787.2 = 1212.8 mm of Hg
Let the rate of diffusion of other gas be rn, then
\frac{r_{n}}{r_{o_{2}}}\sqrt{\frac{32}{79}}
∴ Drop in pressure for the other gas = 787.2 * 32/79
= 501.01 mm of Hg
∴ pressure of the other gas after 74 minutes
= 2000 – 501.01 mm = 1498.99 mm of Hg
Molar ratio = Moles of unknown gas/Moles of O2
= 1498.99/1212.8 = 1.24/1 = 1.24 : 1.
[Partial pressure ∝ mole fraction]
Last Activity: 11 Years ago
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