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Direction (Q. Nos. 13 and 16) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive)
Q. A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.
During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)
During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)
Correct answer is '9'. Can you explain this answer?
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Direction (Q. Nos. 13 and 16) This section contains 4questions. when w...
We know that
pV = nRT
A/Q , p and T are constant;
So n1/n2 = V1/V2
(¼)/(0.25/4) = (10)3/r3
r3 = 750
r ∼ 9
pV = nRT
A/Q , p and T are constant;
So n1/n2 = V1/V2
(¼)/(0.25/4) = (10)3/r3
r3 = 750
r ∼ 9
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Direction (Q. Nos. 13 and 16) This section contains 4questions. when w...
To solve this problem, we can use the ideal gas law and the equation for effusion.
1. Calculate the initial volume of the balloon:
The volume of the balloon can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the balloon.
Given that the radius is 10.0 cm, we can substitute this value into the formula to find the initial volume of the balloon.
V_initial = (4/3)π(10.0 cm)^3
V_initial = (4/3)π(1000 cm^3)
V_initial = (4/3)(3.14)(1000 cm^3)
V_initial ≈ 4186.67 cm^3
2. Calculate the initial number of moles of helium gas:
Using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature, we can rearrange the equation to solve for n.
n = PV/RT
Given that the pressure and temperature are constant, we can substitute the initial volume of the balloon and the mass of helium gas into the equation to find the initial number of moles.
n_initial = (PV_initial)/(RT)
n_initial = (1.00 g)/(4 g/mol)
n_initial = 0.25 mol
3. Calculate the final number of moles of helium gas:
Since 0.25 g of gas effused from the balloon overnight, the final number of moles can be calculated by subtracting the effused mass from the initial mass and dividing by the molar mass of helium.
n_final = (m_initial - m_effused)/(molar mass)
n_final = (1.00 g - 0.25 g)/(4 g/mol)
n_final = 0.1875 mol
4. Calculate the final volume of the balloon:
Using the ideal gas law equation and the final number of moles of helium gas, we can rearrange the equation to solve for the final volume.
V_final = (n_final RT)/P
Since the pressure and temperature are constant, we can substitute the values into the equation to find the final volume.
V_final = (n_final RT_initial)/P
V_final = (0.1875 mol)(0.0821 L/mol K)(273 K)/(1 atm)
V_final ≈ 3.81 L
V_final ≈ 3810 cm^3
5. Calculate the final radius of the balloon:
Using the formula for the volume of a sphere, we can rearrange the equation to solve for the final radius.
V_final = (4/3)πr_final^3
Substituting the final volume into the equation, we can solve for the final radius.
(4/3)πr_final^3 = V_final
(4/3)πr_final^3 = 3810 cm^3
Solving for r_final:
r_final^3 = (3/4)(3810 cm^3)/π
r_final^3 ≈ 2865 cm^3/π
r_final^3 ≈ 912.14
r_final ≈ 9 cm
Therefore, the radius of the balloon the next morning is approximately 9 cm
1. Calculate the initial volume of the balloon:
The volume of the balloon can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the balloon.
Given that the radius is 10.0 cm, we can substitute this value into the formula to find the initial volume of the balloon.
V_initial = (4/3)π(10.0 cm)^3
V_initial = (4/3)π(1000 cm^3)
V_initial = (4/3)(3.14)(1000 cm^3)
V_initial ≈ 4186.67 cm^3
2. Calculate the initial number of moles of helium gas:
Using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature, we can rearrange the equation to solve for n.
n = PV/RT
Given that the pressure and temperature are constant, we can substitute the initial volume of the balloon and the mass of helium gas into the equation to find the initial number of moles.
n_initial = (PV_initial)/(RT)
n_initial = (1.00 g)/(4 g/mol)
n_initial = 0.25 mol
3. Calculate the final number of moles of helium gas:
Since 0.25 g of gas effused from the balloon overnight, the final number of moles can be calculated by subtracting the effused mass from the initial mass and dividing by the molar mass of helium.
n_final = (m_initial - m_effused)/(molar mass)
n_final = (1.00 g - 0.25 g)/(4 g/mol)
n_final = 0.1875 mol
4. Calculate the final volume of the balloon:
Using the ideal gas law equation and the final number of moles of helium gas, we can rearrange the equation to solve for the final volume.
V_final = (n_final RT)/P
Since the pressure and temperature are constant, we can substitute the values into the equation to find the final volume.
V_final = (n_final RT_initial)/P
V_final = (0.1875 mol)(0.0821 L/mol K)(273 K)/(1 atm)
V_final ≈ 3.81 L
V_final ≈ 3810 cm^3
5. Calculate the final radius of the balloon:
Using the formula for the volume of a sphere, we can rearrange the equation to solve for the final radius.
V_final = (4/3)πr_final^3
Substituting the final volume into the equation, we can solve for the final radius.
(4/3)πr_final^3 = V_final
(4/3)πr_final^3 = 3810 cm^3
Solving for r_final:
r_final^3 = (3/4)(3810 cm^3)/π
r_final^3 ≈ 2865 cm^3/π
r_final^3 ≈ 912.14
r_final ≈ 9 cm
Therefore, the radius of the balloon the next morning is approximately 9 cm
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Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer?
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Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer?.
Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer?.
Solutions for Direction (Q. Nos. 13 and 16) This section contains 4questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q.A toy balloon originally held 1.00 g helium gas and had a radius of 10.0 cm.During the night 0.25 g of the gas effused from the balloon. Assuming the ideal gas behaviour under the constant pressure and temperature conditions, what was the radius of the balloon the next morning? (Express result in cm.)Correct answer is '9'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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