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A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :
Correct answer is '6'. Can you explain this answer?
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A spherical balloon of volume 5 litre is to be filled up with H2 at NT...
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A spherical balloon of volume 5 litre is to be filled up with H2 at NT...
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
First, let's calculate the number of moles of H2 in the cylinder using the given information:
P1 = 6 atm (pressure in the cylinder)
V1 = 6 L (volume of the cylinder)
n1 = ? (number of moles of H2 in the cylinder)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T1 = 0°C = 273.15 K (temperature in Kelvin)
Using the ideal gas law equation:
P1V1 = n1RT1
6 atm * 6 L = n1 * 0.0821 L·atm/(mol·K) * 273.15 K
36 atm·L = n1 * 22.414 L·atm/(mol·K)
n1 = (36 atm·L) / (22.414 L·atm/(mol·K))
n1 = 1.607 mol
So, there are 1.607 moles of H2 in the cylinder.
Now, let's calculate the number of moles of H2 needed to fill the balloon using the ideal gas law equation:
P2 = 1 atm (pressure in the balloon at NTP)
V2 = 5 L (volume of the balloon)
n2 = ? (number of moles of H2 needed to fill the balloon)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T2 = 0°C = 273.15 K (temperature in Kelvin)
Using the ideal gas law equation:
P2V2 = n2RT2
1 atm * 5 L = n2 * 0.0821 L·atm/(mol·K) * 273.15 K
5 atm·L = n2 * 22.414 L·atm/(mol·K)
n2 = (5 atm·L) / (22.414 L·atm/(mol·K))
n2 = 1.118 mol
So, 1.118 moles of H2 are needed to fill the balloon.
Since we have 1.607 moles of H2 in the cylinder and we need only 1.118 moles to fill the balloon, we have an excess of 1.607 - 1.118 = 0.489 moles of H2.
Therefore, after filling up the balloon, there will still be 0.489 moles of H2 remaining in the cylinder.
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
First, let's calculate the number of moles of H2 in the cylinder using the given information:
P1 = 6 atm (pressure in the cylinder)
V1 = 6 L (volume of the cylinder)
n1 = ? (number of moles of H2 in the cylinder)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T1 = 0°C = 273.15 K (temperature in Kelvin)
Using the ideal gas law equation:
P1V1 = n1RT1
6 atm * 6 L = n1 * 0.0821 L·atm/(mol·K) * 273.15 K
36 atm·L = n1 * 22.414 L·atm/(mol·K)
n1 = (36 atm·L) / (22.414 L·atm/(mol·K))
n1 = 1.607 mol
So, there are 1.607 moles of H2 in the cylinder.
Now, let's calculate the number of moles of H2 needed to fill the balloon using the ideal gas law equation:
P2 = 1 atm (pressure in the balloon at NTP)
V2 = 5 L (volume of the balloon)
n2 = ? (number of moles of H2 needed to fill the balloon)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T2 = 0°C = 273.15 K (temperature in Kelvin)
Using the ideal gas law equation:
P2V2 = n2RT2
1 atm * 5 L = n2 * 0.0821 L·atm/(mol·K) * 273.15 K
5 atm·L = n2 * 22.414 L·atm/(mol·K)
n2 = (5 atm·L) / (22.414 L·atm/(mol·K))
n2 = 1.118 mol
So, 1.118 moles of H2 are needed to fill the balloon.
Since we have 1.607 moles of H2 in the cylinder and we need only 1.118 moles to fill the balloon, we have an excess of 1.607 - 1.118 = 0.489 moles of H2.
Therefore, after filling up the balloon, there will still be 0.489 moles of H2 remaining in the cylinder.
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A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer?
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A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer?.
A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical balloon of volume 5 litre is to be filled up with H2 at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is :Correct answer is '6'. Can you explain this answer?.
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