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A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)
[JEE Main 2013]
- a)q = + 208 J, W = - 208
- b)q = - 208 J, W = - 208 J
- c)q = - 208 J, W = + 208 J
- d)q = + 208 J, W = + 208 J
Correct answer is option 'A'. Can you explain this answer?
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A piston filled with 0.04 mole of an ideal gas expands reversibly from...
Q = + 208 J (as it absorb heat)
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A piston filled with 0.04 mole of an ideal gas expands reversibly from...
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = gas constant
T = temperature
Since the temperature is constant, we can rewrite the equation as:
(P1 * V1) / n1 = (P2 * V2) / n2
Where:
P1 = initial pressure
V1 = initial volume
n1 = initial moles of gas
P2 = final pressure
V2 = final volume
n2 = final moles of gas
We can rearrange the equation to solve for the final pressure:
P2 = (P1 * V1 * n2) / (V2 * n1)
Given:
P1 = unknown
V1 = 50.0 mL = 0.05 L
n1 = 0.04 mole
V2 = 375.0 mL = 0.375 L
n2 = 0.04 mole
R = gas constant = 0.0821 L·atm/(mol·K)
T = 37.0°C = 37.0 + 273.15 = 310.15 K
Plugging in the values:
P2 = (P1 * 0.05 L * 0.04 mole) / (0.375 L * 0.04 mole)
P2 = (P1 * 0.002 L*mol) / (0.015 L*mol)
P2 = (P1 * 0.002) / 0.015
P2 = (P1 * 0.1333)
Now we need to solve for P1. We can use the ideal gas law equation with the initial conditions:
(P1 * 0.05 L) / (0.04 mole) = n1 * R * T
(P1 * 0.05 L) / (0.04 mole) = 0.04 mole * 0.0821 L·atm/(mol·K) * 310.15 K
(P1 * 0.05 L) / (0.04 mole) = 1.01 L·atm
Simplifying the equation:
P1 = (1.01 L·atm * 0.04 mole) / 0.05 L
P1 = 0.808 L·atm
Now we can substitute this value back into the equation for P2:
P2 = (0.808 L·atm * 0.002) / 0.015
P2 = 0.107 L·atm
Therefore, the final pressure of the gas is approximately 0.107 L·atm.
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = gas constant
T = temperature
Since the temperature is constant, we can rewrite the equation as:
(P1 * V1) / n1 = (P2 * V2) / n2
Where:
P1 = initial pressure
V1 = initial volume
n1 = initial moles of gas
P2 = final pressure
V2 = final volume
n2 = final moles of gas
We can rearrange the equation to solve for the final pressure:
P2 = (P1 * V1 * n2) / (V2 * n1)
Given:
P1 = unknown
V1 = 50.0 mL = 0.05 L
n1 = 0.04 mole
V2 = 375.0 mL = 0.375 L
n2 = 0.04 mole
R = gas constant = 0.0821 L·atm/(mol·K)
T = 37.0°C = 37.0 + 273.15 = 310.15 K
Plugging in the values:
P2 = (P1 * 0.05 L * 0.04 mole) / (0.375 L * 0.04 mole)
P2 = (P1 * 0.002 L*mol) / (0.015 L*mol)
P2 = (P1 * 0.002) / 0.015
P2 = (P1 * 0.1333)
Now we need to solve for P1. We can use the ideal gas law equation with the initial conditions:
(P1 * 0.05 L) / (0.04 mole) = n1 * R * T
(P1 * 0.05 L) / (0.04 mole) = 0.04 mole * 0.0821 L·atm/(mol·K) * 310.15 K
(P1 * 0.05 L) / (0.04 mole) = 1.01 L·atm
Simplifying the equation:
P1 = (1.01 L·atm * 0.04 mole) / 0.05 L
P1 = 0.808 L·atm
Now we can substitute this value back into the equation for P2:
P2 = (0.808 L·atm * 0.002) / 0.015
P2 = 0.107 L·atm
Therefore, the final pressure of the gas is approximately 0.107 L·atm.
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A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer?
Question Description
A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. 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 A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. 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 A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer?.
A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. 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 A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. 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 A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer?.
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A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)[JEE Main 2013]a)q = + 208 J, W = - 208b)q = - 208 J, W = - 208 Jc)q = - 208 J, W = + 208 Jd)q = + 208 J, W = + 208 JCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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