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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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    The atmospheric lapse rateFor small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantityTP(1 – λ)/λ has a uniform value through the layers of troposphere.(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)Q. If behaviour of the mixing of parcels of air is approximately assumed to be adiabatic then lapse rate can be expressed as

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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?.
    Solutions 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? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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