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25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, are
  • a)
    -1036.194 cal and 12,464.886 cal
  • b)
    921.4 cal and 11,074 cal
  • c)
    1029.4 cal and 12,470.6 cal
  • d)
    1129.3 cal and 10,207 cal
Correct answer is option 'A'. Can you explain this answer?
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25 grams of water vaporises at 373 K against a constant external press...
Given data:
- Mass of water vapor = 25 grams
- Temperature = 373 K
- External pressure = 1 atm
- Molar enthalpy of vaporisation = 9.72 kcal/mol

Calculations:
1. Calculate the number of moles of water vapor:
Number of moles = Mass / Molar mass
Molar mass of water = 18 g/mol
Number of moles = 25 g / 18 g/mol = 1.389 moles
2. Calculate the work done against the atmosphere:
Work done = -PΔV
ΔV = nRT/P (from ideal gas law)
ΔV = 1.389 * 0.0821 * 373 / 1 = 43.96 L
Work done = -1 atm * 43.96 L = -43.96 L.atm
Convert work to calories: 1 L.atm = 101.3 J = 0.241 cal
Work done = -43.96 * 0.241 cal = -10.6 cal
3. Calculate the change in internal energy:
Change in internal energy = q + w
q = nΔH
q = 1.389 * 9.72 kcal = 13.494 kcal
Change in internal energy = 13.494 - 10.6 = 2.894 kcal = 2894 cal

Final answer:
- Work done against the atmosphere = -10.6 cal = -1036.194 cal (rounded off)
- Change in internal energy = 2894 cal
Therefore, option 'A' is the correct answer.
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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. The value of theoretical lapse rate on the earth is (use g = 9.8 m/s2 ; R = 8.3 J/mol-k and M = 29 g/mol)

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

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. Mechanical equilibrium of the atmosphere requires that the pressure decreases with altitude according to . Assuming free fall acceleration to be uniform, then lapse rate is given by

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25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer?
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25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. 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 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer?.
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