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10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter?
- a)31842 kg
- b)58.55 kg
- c)342.58 kg
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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10 mole of ideal gas expand isothermally and reversibly from a pressur...
Given:
- Number of moles of ideal gas, n = 10 mol
- Initial pressure, P₁ = 10 atm
- Final pressure, P₂ = 1 atm
- Temperature, T = 300 K
- Height, h = 100 m
Assumption:
- The gas behaves ideally.
- The expansion is isothermal and reversible.
- The gravitational acceleration, g = 9.8 m/s².
Explanation:
Step 1: Calculate the work done by the gas during expansion.
The work done by an ideal gas during expansion is given by the equation:
W = -nRT ln(V₂/V₁)
Where:
- W is the work done by the gas on the surroundings
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature of the gas in Kelvin
- V₁ and V₂ are the initial and final volumes of the gas, respectively
Since the expansion is isothermal and reversible, we can use the ideal gas equation to relate the initial and final volumes:
P₁V₁ = P₂V₂
Rearranging the equation, we get:
V₂/V₁ = P₁/P₂
Substituting this into the equation for work done:
W = -nRT ln(P₁/P₂)
W = -10 * 8.314 * 300 * ln(10/1)
W ≈ -10 * 8.314 * 300 * 2.3026
W ≈ -58,555 J
Step 2: Calculate the potential energy gained by lifting the object.
The potential energy gained by lifting the object is given by the equation:
PE = mgh
Where:
- PE is the potential energy gained
- m is the mass of the object
- g is the gravitational acceleration
- h is the height lifted
Rearranging the equation, we can solve for the mass of the object:
m = PE / (gh)
Substituting the given values:
m = (-58,555 J) / (10 kg) * (9.8 m/s²) * (100 m)
m ≈ -58.55 kg
Step 3: Conclusion
The largest mass that can be lifted through a height of 100 meters is approximately 58.55 kg. Therefore, the correct answer is option B.
- Number of moles of ideal gas, n = 10 mol
- Initial pressure, P₁ = 10 atm
- Final pressure, P₂ = 1 atm
- Temperature, T = 300 K
- Height, h = 100 m
Assumption:
- The gas behaves ideally.
- The expansion is isothermal and reversible.
- The gravitational acceleration, g = 9.8 m/s².
Explanation:
Step 1: Calculate the work done by the gas during expansion.
The work done by an ideal gas during expansion is given by the equation:
W = -nRT ln(V₂/V₁)
Where:
- W is the work done by the gas on the surroundings
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature of the gas in Kelvin
- V₁ and V₂ are the initial and final volumes of the gas, respectively
Since the expansion is isothermal and reversible, we can use the ideal gas equation to relate the initial and final volumes:
P₁V₁ = P₂V₂
Rearranging the equation, we get:
V₂/V₁ = P₁/P₂
Substituting this into the equation for work done:
W = -nRT ln(P₁/P₂)
W = -10 * 8.314 * 300 * ln(10/1)
W ≈ -10 * 8.314 * 300 * 2.3026
W ≈ -58,555 J
Step 2: Calculate the potential energy gained by lifting the object.
The potential energy gained by lifting the object is given by the equation:
PE = mgh
Where:
- PE is the potential energy gained
- m is the mass of the object
- g is the gravitational acceleration
- h is the height lifted
Rearranging the equation, we can solve for the mass of the object:
m = PE / (gh)
Substituting the given values:
m = (-58,555 J) / (10 kg) * (9.8 m/s²) * (100 m)
m ≈ -58.55 kg
Step 3: Conclusion
The largest mass that can be lifted through a height of 100 meters is approximately 58.55 kg. Therefore, the correct answer is option B.
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10 mole of ideal gas expand isothermally and reversibly from a pressur...
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10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter?a)31842 kgb)58.55 kgc)342.58 kgd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter?a)31842 kgb)58.55 kgc)342.58 kgd)None of theseCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter?a)31842 kgb)58.55 kgc)342.58 kgd)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. What is the largest mass which can lifted through a height of 100 meter?a)31842 kgb)58.55 kgc)342.58 kgd)None of theseCorrect answer is option 'B'. Can you explain this answer?.
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