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Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape ?
- a)1/2
- b)1/8
- c)1/4
- d)3/8
Correct answer is option 'B'. Can you explain this answer?
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Equal moles of hydrogen and oxygen gases are placed in a container wit...
we have given,
a number of moles of hydrogen and (nH2) that of oxygen are(nO2) not equal.
a number of moles of hydrogen and (nH2) that of oxygen are(nO2) not equal.
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Equal moles of hydrogen and oxygen gases are placed in a container wit...
Given: Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape.
To find: What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?
Let's assume that n moles of both hydrogen and oxygen are present in the container.
According to Graham's Law of Diffusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Thus, the rate of effusion of hydrogen gas will be √(32/2) = √16 = 4 times faster than oxygen gas.
Therefore, the time required for half of the hydrogen gas to escape will also be 4 times faster than oxygen gas.
Let's assume that it takes t seconds for half of the hydrogen gas to escape.
So, the time required for half of the oxygen gas to escape will be 4t seconds.
Now, we know that the fraction of a gas that escapes in a given time is directly proportional to its rate of effusion.
Thus, the fraction of oxygen gas that escapes in time t will be proportional to the rate of effusion of oxygen gas, which is √(32/n).
Similarly, the fraction of oxygen gas that escapes in time 4t will be proportional to the rate of effusion of oxygen gas, which is √(32/n).
Therefore, the required fraction of oxygen gas that escapes in time t/t+4t or 1/5 is given by the ratio of the two fractions:
Fraction of oxygen gas that escapes in time t/t+4t = (√(32/n))/(√(32/n) + 4√(32/n)) = 1/8.
Hence, the correct answer is option (b) 1/8.
To find: What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?
Let's assume that n moles of both hydrogen and oxygen are present in the container.
According to Graham's Law of Diffusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Thus, the rate of effusion of hydrogen gas will be √(32/2) = √16 = 4 times faster than oxygen gas.
Therefore, the time required for half of the hydrogen gas to escape will also be 4 times faster than oxygen gas.
Let's assume that it takes t seconds for half of the hydrogen gas to escape.
So, the time required for half of the oxygen gas to escape will be 4t seconds.
Now, we know that the fraction of a gas that escapes in a given time is directly proportional to its rate of effusion.
Thus, the fraction of oxygen gas that escapes in time t will be proportional to the rate of effusion of oxygen gas, which is √(32/n).
Similarly, the fraction of oxygen gas that escapes in time 4t will be proportional to the rate of effusion of oxygen gas, which is √(32/n).
Therefore, the required fraction of oxygen gas that escapes in time t/t+4t or 1/5 is given by the ratio of the two fractions:
Fraction of oxygen gas that escapes in time t/t+4t = (√(32/n))/(√(32/n) + 4√(32/n)) = 1/8.
Hence, the correct answer is option (b) 1/8.
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Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through whichboth can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogento escape ?a)1/2b)1/8c)1/4d)3/8Correct answer is option 'B'. Can you explain this answer?
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