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For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___
(Important - Enter only the numerical value in the answer)
Correct answer is '2443.25'. Can you explain this answer?
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For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressur...
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Estimating the Enthalpy of Evaporation using the Clausius-Clapeyron Equation
The Clausius-Clapeyron equation relates the saturation pressure and temperature of a substance to its enthalpy of evaporation. It is given by:
ln(ps2/ps1) = ΔHvap/R * (1/T1 - 1/T2)
where ps1 and ps2 are the saturation pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of evaporation, and R is the gas constant.
Given information:
- dps/dTs = 0.189 kPa/K
- Specific volume of dry saturated vapour (vapour phase) = 43.38 m3/kg
We need to estimate the enthalpy of evaporation of water at 25°C.
Step 1: Convert temperature to Kelvin
T1 = 25°C = 298 K
Step 2: Calculate the saturation pressure at T1 using the given information
ps1 = ps(T1)
To do this, we can integrate the given rate of change of saturation pressure with respect to temperature:
∫dps = ∫(0.189 kPa/K)dTs
Integrating both sides gives:
ps = 0.189T + C
To find the constant C, we can use the fact that ps = 0.61165 kPa at the boiling point of water (100°C = 373.15 K):
0.61165 = 0.189(373.15) + C
Solving for C gives:
C = 0.61165 - 0.189(373.15)
Substituting the value of C back into the equation for ps, we can calculate the saturation pressure at T1:
ps1 = 0.189T1 + (0.61165 - 0.189(373.15))
Step 3: Convert the specific volume of dry saturated vapour to specific volume of liquid
Given that the specific volume of liquid is negligible compared to that of vapour, we can assume that the specific volume of liquid is zero.
Step 4: Use the Clausius-Clapeyron equation to estimate the enthalpy of evaporation
We can rearrange the Clausius-Clapeyron equation to solve for ΔHvap:
ΔHvap = R * (1/T1 - 1/T2) * ln(ps2/ps1)
Substituting the known values into the equation, we get:
ΔHvap = R * (1/298 - 1/T2) * ln(ps2/ps1)
Step 5: Calculate the enthalpy of evaporation at T2 = T1
ΔHvap = R * (1/298 - 1/298) * ln(ps1/ps1)
ΔHvap = 0
Therefore, the enthalpy of evaporation at 25°C is zero.
The correct answer provided, 2443.25 kJ/kg, does not match the calculations based on the given information. It is possible that there is additional information or an error in the given problem statement.
The Clausius-Clapeyron equation relates the saturation pressure and temperature of a substance to its enthalpy of evaporation. It is given by:
ln(ps2/ps1) = ΔHvap/R * (1/T1 - 1/T2)
where ps1 and ps2 are the saturation pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of evaporation, and R is the gas constant.
Given information:
- dps/dTs = 0.189 kPa/K
- Specific volume of dry saturated vapour (vapour phase) = 43.38 m3/kg
We need to estimate the enthalpy of evaporation of water at 25°C.
Step 1: Convert temperature to Kelvin
T1 = 25°C = 298 K
Step 2: Calculate the saturation pressure at T1 using the given information
ps1 = ps(T1)
To do this, we can integrate the given rate of change of saturation pressure with respect to temperature:
∫dps = ∫(0.189 kPa/K)dTs
Integrating both sides gives:
ps = 0.189T + C
To find the constant C, we can use the fact that ps = 0.61165 kPa at the boiling point of water (100°C = 373.15 K):
0.61165 = 0.189(373.15) + C
Solving for C gives:
C = 0.61165 - 0.189(373.15)
Substituting the value of C back into the equation for ps, we can calculate the saturation pressure at T1:
ps1 = 0.189T1 + (0.61165 - 0.189(373.15))
Step 3: Convert the specific volume of dry saturated vapour to specific volume of liquid
Given that the specific volume of liquid is negligible compared to that of vapour, we can assume that the specific volume of liquid is zero.
Step 4: Use the Clausius-Clapeyron equation to estimate the enthalpy of evaporation
We can rearrange the Clausius-Clapeyron equation to solve for ΔHvap:
ΔHvap = R * (1/T1 - 1/T2) * ln(ps2/ps1)
Substituting the known values into the equation, we get:
ΔHvap = R * (1/298 - 1/T2) * ln(ps2/ps1)
Step 5: Calculate the enthalpy of evaporation at T2 = T1
ΔHvap = R * (1/298 - 1/298) * ln(ps1/ps1)
ΔHvap = 0
Therefore, the enthalpy of evaporation at 25°C is zero.
The correct answer provided, 2443.25 kJ/kg, does not match the calculations based on the given information. It is possible that there is additional information or an error in the given problem statement.
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For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer?
Question Description
For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer?.
For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer?.
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For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer?, a detailed solution for For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? has been provided alongside types of For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice For water at 25oC, dps/dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparson with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25oC (in kJ/kg) is ___(Important - Enter only the numerical value in the answer)Correct answer is '2443.25'. Can you explain this answer? tests, examples and also practice GATE tests.
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