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A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.
Correct answer is '55.00'. Can you explain this answer?
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A cylinder supports a piston of mass 5 kg5 kg and cross-sectional are...
During heating, the process was isobaric, so, ΔW1 = PΔV
And internal energy, ΔU1 = nCvΔT = 50nCv
Thus, heat transfer, ΔQ1 = 50nCv + 55
During cooling, the process is isochoric, so, ΔW2 = 0 and ΔU2 = −50nCv
Thus, heat loss = ΔQ2 = −(ΔU2 + ΔW2) = 50nCv
∴ Heat supplied −- Heat rejected = 50nCv + 55 − 50nCv = 55 J
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A cylinder supports a piston of mass 5 kg5 kg and cross-sectional are...
Given data:
- Mass of piston = 5 kg
- Cross-sectional area of piston = 5×10^−3 m^2
- Initial temperature = 27°C
- Final temperature = 77°C
- Piston displacement = 0.1 m
- Atmospheric pressure = 1 atm
Calculating work done during heating:
- Work done = Pressure × Change in volume
- Change in volume = Area of piston × Piston displacement
- Change in volume = 5×10^−3 m^2 × 0.1 m = 0.0005 m^3
- Pressure = Atmospheric pressure = 1 atm = 1.01325 × 10^5 Pa
- Work done = 1.01325 × 10^5 Pa × 0.0005 m = 50.6625 J
Calculating heat supplied during heating:
- Heat supplied = Work done + Change in internal energy
- Change in internal energy = nCvΔT (for ideal gas)
- n = number of moles of gas = PV/RT
- P = Pressure = 1 atm = 1.01325 × 10^5 Pa
- V = Volume = Area of piston × Piston displacement = 5×10^−3 m^2 × 0.1 m = 0.0005 m^3
- R = Ideal gas constant = 8.314 J/mol·K
- T = Average temperature = (Initial temperature + Final temperature) / 2 = (27°C + 77°C) / 2 = 52°C = 325 K
- Cv = Specific heat at constant volume = 3/2R = 3/2 × 8.314 J/mol·K = 12.471 J/mol·K
- Change in internal energy = nCvΔT = PV/RT × 12.471 J/mol·K × 325 K = 1.01325 × 10^5 Pa × 0.0005 m^3 / (8.314 J/mol·K × 325 K) × 12.471 J/mol·K × 325 K = 24.942 J
- Heat supplied = 50.6625 J + 24.942 J = 75.6045 J
Calculating heat rejected during cooling:
- Since the process is reversible, the heat rejected during cooling will be equal to the heat supplied during heating
- Heat rejected = 75.6045 J
Calculating the difference in heat supplied and heat rejected:
- Difference = Heat supplied - Heat rejected = 75.6045 J - 75.6045 J = 0 J
Therefore, the difference in heat supplied during heating and heat rejected during cooling is 0 J.
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A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. Can you explain this answer?
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A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. 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 cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. 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 cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. Can you explain this answer?.
A cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. 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 cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. 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 cylinder supports a piston of mass 5 kg5 kg and cross-sectional area 5×10−3 m2 enclosing a gas at 27°C. The gas is slowly heated to 77°C such that the piston rises by 0.1 m. The piston is now clamped at this new position and the gas is cooled down to its initial temperature. If atmospheric pressure is 1 atm, then the difference in heat supplied during heating and heat rejected during cooling is _____J.Correct answer is '55.00'. Can you explain this answer?.
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