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A 4 : 1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially.
Correct answer is '8 : 1'. Can you explain this answer?
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A 4 : 1 molar mixture of He and CH4 is contained in a vessel at 20 bar...
Total moles = 4+1 =5, Total pressure = 20 bars
So, partial pressure of He = 4/5 x 20 = 16 bar
Or, partial pressure of CH4 = 20-16 = 4 bar
Now, molar mass of He = 4, molar mass of CH4 = 16
So, partial pressure of He = 4/5 x 20 = 16 bar
Or, partial pressure of CH4 = 20-16 = 4 bar
Now, molar mass of He = 4, molar mass of CH4 = 16
So, ratio of rate of effusion of He (rHe) and rate of effusion of methane (rCH4) will be:
Hence, as the rate of effusion of He is 8 times higher than CH4 so, initial mixture must contain the molar ratio of He: CH4 to be 8:1.
Hence, as the rate of effusion of He is 8 times higher than CH4 so, initial mixture must contain the molar ratio of He: CH4 to be 8:1.
Most Upvoted Answer
A 4 : 1 molar mixture of He and CH4 is contained in a vessel at 20 bar...
Solution:
Effusion is the process of a gas escaping through a small hole into a vacuum. According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The rate of effusion of He and CH4 can be determined by using the following equation:
Rate of effusion of gas A/Rate of effusion of gas B = √(Molar mass of gas B/Molar mass of gas A)
Let us assume that the rate of effusion of He is x and the rate of effusion of CH4 is y.
x/y = √(Molar mass of CH4/Molar mass of He)
x/y = √(16/4) (Molar mass of CH4 = 16 and Molar mass of He = 4)
x/y = 2
x = 2y
Therefore, the ratio of the rates of effusion of He and CH4 is 2 : 1.
Since the pressure of the gas mixture is 20 bar, the partial pressure of He in the mixture is:
Partial pressure of He = (4/5) x 20 bar = 16 bar
Similarly, the partial pressure of CH4 in the mixture is:
Partial pressure of CH4 = (1/5) x 20 bar = 4 bar
The composition of the mixture effusing out initially can be determined by dividing the partial pressures of the gases in the mixture by the total pressure of the mixture.
The composition of the mixture effusing out initially is:
(16 bar / 20 bar) : (4 bar / 20 bar) = 8 : 1
Therefore, the correct answer is '8 : 1'.
Effusion is the process of a gas escaping through a small hole into a vacuum. According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The rate of effusion of He and CH4 can be determined by using the following equation:
Rate of effusion of gas A/Rate of effusion of gas B = √(Molar mass of gas B/Molar mass of gas A)
Let us assume that the rate of effusion of He is x and the rate of effusion of CH4 is y.
x/y = √(Molar mass of CH4/Molar mass of He)
x/y = √(16/4) (Molar mass of CH4 = 16 and Molar mass of He = 4)
x/y = 2
x = 2y
Therefore, the ratio of the rates of effusion of He and CH4 is 2 : 1.
Since the pressure of the gas mixture is 20 bar, the partial pressure of He in the mixture is:
Partial pressure of He = (4/5) x 20 bar = 16 bar
Similarly, the partial pressure of CH4 in the mixture is:
Partial pressure of CH4 = (1/5) x 20 bar = 4 bar
The composition of the mixture effusing out initially can be determined by dividing the partial pressures of the gases in the mixture by the total pressure of the mixture.
The composition of the mixture effusing out initially is:
(16 bar / 20 bar) : (4 bar / 20 bar) = 8 : 1
Therefore, the correct answer is '8 : 1'.
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A 4 : 1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially.Correct answer is '8 : 1'. Can you explain this answer?
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