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The pressure of bulb dropped from 2000mmhg to 1500mmhg in 47 min, when a contained o2 gas is leaved through a small hole, the bulb was then evaculated and the mixture of o2 and another gas (m=79)in the mole ratio 1 :1 at total pressue of 4000mmhg was introduced. Find the molar ratio of two gases of the bulb after 74 min ?
Accepted Answer
This problem involves the concept of effusion and Graham's Law. Here's how to solve it:
1. Effusion Rate and Graham's Law:
The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham's Law).
The pressure drop in the bulb is directly related to the amount of O2 that escaped, allowing us to calculate the relative effusion rates.
2. Calculating the Effusion Rate Ratio:
Let's denote the effusion rate of O2 as 'r1' and the effusion rate of the unknown gas (m=79) as 'r2'.
The ratio of effusion rates (r1/r2) is equal to the square root of the inverse ratio of their molar masses: r1/r2 = √(M2/M1) = √(79/32)
3. Calculating the Amount of O2 Escaped:
The pressure dropped from 2000 mmHg to 1500 mmHg, indicating a 500 mmHg decrease. This pressure decrease corresponds to the amount of O2 that escaped.
4. Calculating the Initial Amount of O2 and the Unknown Gas:
After evacuation and filling, the total pressure is 4000 mmHg. Since the mole ratio of O2 and the unknown gas is 1:1, the partial pressure of each gas is 2000 mmHg.
5. Calculating the Amount of Unknown Gas That Effused:
Using the effusion rate ratio (r1/r2) calculated earlier, we can determine the amount of the unknown gas that escaped during the 74 minutes.
6. Calculating the Final Molar Ratio:
Subtract the amount of O2 and the unknown gas that escaped from their initial amounts to find the final amount of each gas in the bulb.
Calculate the final molar ratio of the two gases.
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