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The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.
Correct answer is '1 : 1.18'. Can you explain this answer?
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The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 m...
I assume that the bulb was placed in a large evacuated space - otherwise, the pressure outside the bulb would rise and slow down the (net) rate of leaking (because gas could leak back in). I also assume that the gases were at the same temperature, because this affects the average kinetic energy of molecules, and hence their speed. Also that the temperature of gas in the bulb was held constant as the gas leaked out.
According to Dalton's Law, the Partial Pressures of the separate gases in a mixture are in their mole ratio, which is 1:1 here, so in the mixture the O2 and 2nd gas each have a PP of 2000mmHg (4000mmHg total) initially.
The rate of diffusion (measured as rate of decrease in pressure) is proportional to pressure dp/dt = -Kp. This has solution p(t) = p(0).exp(-Kt) where K is a diffusion constant which depends on the molecular weight and temperature.
For O2 (experiment 1), p(0) = 2000mmHg and p(47) = p(0).exp(-47K) so
-47K = ln(p(47)/p(0)) = ln(1500/2000) = ln(0.75) = -0.287682072
K = 0.006120895.
In experiment 2 (mixed gas), if the time is 47 mins, the final PP of O2 will be 1500mmHg as before; if the time is 74 mins, the final PP is p(74) = p(0).exp(-74K) = 2000mmHg x 0.63575231 = 1271.5mmHg.
According to Graham's Law, the rate of diffusion is inversely proportional to square root of molecular weight, which is 32 for O2, 79 for unknown gas. So the value of K for the unknown gas is
K' = K x √(32/79) = 0.003895618.
So for unknown gas, if time is 47 mins, p(47) = p(0).exp(-47K') = 2000mmHg x 0.832689834 =1665.4mmHg.
If time is 74 mins, p(74) = 2000mmHg x exp(-74K') = 2000 x 0.749554877 = 1499.1 mmHg.
The final mole ratio is the same as the ratio of partial pressures:
if t=47 mins: 1500:1655.4 = approx. 10:11,
if t=74 mins: 1271.5:1499.1.
Most Upvoted Answer
The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 m...
Given:
- Initial pressure in the bulb = 2000 mmHg
- Final pressure in the bulb = 1500 mmHg
- Time taken for the pressure to drop = 47 min
- Total pressure of the gas mixture introduced = 4000 mmHg
- Molecular weight of the other gas = 79
To Find:
- Molar ratio of the two gases remaining in the bulb after 74 min.
Assumptions:
- The volume of the bulb remains constant throughout the process.
- The ideal gas equation is applicable.
Solution:
Step 1: Determining the initial moles of oxygen gas (O2) in the bulb
- We can use the ideal gas equation to calculate the initial moles of oxygen gas:
PV = nRT
n = PV/RT
- Since the volume (V), temperature (T), and the gas constant (R) remain constant, we can write:
n ∝ P
- Therefore, the initial moles of oxygen gas can be calculated as:
n1 = (2000/760) * n1'
- Here, n1' represents the moles of oxygen gas at 760 mmHg (standard pressure).
n1' = PV/RT = (760/760) * n1'
- Thus, n1 = n1'
Step 2: Determining the final moles of oxygen gas in the bulb
- Using the same approach as in Step 1, we can calculate the final moles of oxygen gas:
n2 = (1500/760) * n2'
- Here, n2' represents the moles of oxygen gas at 760 mmHg (standard pressure).
n2' = PV/RT = (760/760) * n2'
- Thus, n2 = n2'
Step 3: Determining the moles of the other gas introduced
- Since the total pressure of the gas mixture introduced is 4000 mmHg, the partial pressure of oxygen gas is:
P1 = (n1/(n1 + n2)) * 4000
- The partial pressure of the other gas is:
P2 = (n2/(n1 + n2)) * 4000
Step 4: Determining the moles of the other gas remaining after 74 min
- The moles of oxygen gas remaining can be calculated using the following equation:
n2' = n2 * (1500/2000) * (74/47)
- The total moles of gas remaining can be calculated as:
n_total = n1 + n2'
- The moles of the other gas remaining can be calculated as:
n2'_other = (n_total - n2') * (P2/4000)
- The molar ratio of the two gases remaining can be calculated as:
molar_ratio = n2'_other / n2'
Step 5: Calculation
- Substitute the given values into the above equations and calculate the molar ratio:
n1 = (2000/760) * n1' (Equation 1)
n2 = (1500/760)
- Initial pressure in the bulb = 2000 mmHg
- Final pressure in the bulb = 1500 mmHg
- Time taken for the pressure to drop = 47 min
- Total pressure of the gas mixture introduced = 4000 mmHg
- Molecular weight of the other gas = 79
To Find:
- Molar ratio of the two gases remaining in the bulb after 74 min.
Assumptions:
- The volume of the bulb remains constant throughout the process.
- The ideal gas equation is applicable.
Solution:
Step 1: Determining the initial moles of oxygen gas (O2) in the bulb
- We can use the ideal gas equation to calculate the initial moles of oxygen gas:
PV = nRT
n = PV/RT
- Since the volume (V), temperature (T), and the gas constant (R) remain constant, we can write:
n ∝ P
- Therefore, the initial moles of oxygen gas can be calculated as:
n1 = (2000/760) * n1'
- Here, n1' represents the moles of oxygen gas at 760 mmHg (standard pressure).
n1' = PV/RT = (760/760) * n1'
- Thus, n1 = n1'
Step 2: Determining the final moles of oxygen gas in the bulb
- Using the same approach as in Step 1, we can calculate the final moles of oxygen gas:
n2 = (1500/760) * n2'
- Here, n2' represents the moles of oxygen gas at 760 mmHg (standard pressure).
n2' = PV/RT = (760/760) * n2'
- Thus, n2 = n2'
Step 3: Determining the moles of the other gas introduced
- Since the total pressure of the gas mixture introduced is 4000 mmHg, the partial pressure of oxygen gas is:
P1 = (n1/(n1 + n2)) * 4000
- The partial pressure of the other gas is:
P2 = (n2/(n1 + n2)) * 4000
Step 4: Determining the moles of the other gas remaining after 74 min
- The moles of oxygen gas remaining can be calculated using the following equation:
n2' = n2 * (1500/2000) * (74/47)
- The total moles of gas remaining can be calculated as:
n_total = n1 + n2'
- The moles of the other gas remaining can be calculated as:
n2'_other = (n_total - n2') * (P2/4000)
- The molar ratio of the two gases remaining can be calculated as:
molar_ratio = n2'_other / n2'
Step 5: Calculation
- Substitute the given values into the above equations and calculate the molar ratio:
n1 = (2000/760) * n1' (Equation 1)
n2 = (1500/760)
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The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer?
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The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer?.
The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 min 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 min.Correct answer is '1 : 1.18'. Can you explain this answer?.
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