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The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5
Correct answer is '136.2'. Can you explain this answer?
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The change in specific volume when 1 kg of water freezes is 91 x 10-6 ...
136.2
From Maxwell’s relation
As given,
T =273 K
It means the pressure under which the ice would freeze = 1 + 135.2 = 136.2 atm.
From Maxwell’s relation
As given,
T =273 K
It means the pressure under which the ice would freeze = 1 + 135.2 = 136.2 atm.
Most Upvoted Answer
The change in specific volume when 1 kg of water freezes is 91 x 10-6 ...
We can use the Clapeyron equation to relate the change in pressure and temperature during a phase change:
dp/dT = ΔH / TΔV
where dp/dT is the rate of change of pressure with respect to temperature, ΔH is the latent heat of the phase change, T is the temperature, and ΔV is the change in specific volume.
At the freezing point of water (273 K), we can solve for the pressure at which ice would form:
dp/dT = (3.66 * 10^5 J/kg) / (273 K * 91 * 10^-6 m^3/kg) = 15.8 MPa/K
To find the pressure at which ice would form, we need to integrate this equation from the initial pressure (i.e. atmospheric pressure) to the pressure at which ice forms. We can assume that the initial specific volume of water is 0.001 m^3/kg (since water is essentially incompressible at atmospheric pressure).
∫dp = ∫(15.8 MPa/K dT) from 273 K to T_f
p_f - 101.3 kPa = 15.8 MPa/K * (T_f - 273 K)
Solving for the pressure at which ice would form (p_f):
p_f = 101.3 kPa + 15.8 MPa/K * (273 K - T_f)
We can use the fact that the change in specific volume during freezing is 91 x 10^-6 m^3/kg to find the final specific volume of ice:
v_f = v_i + Δv = 0.001 m^3/kg + 91 x 10^-6 m^3/kg = 0.001091 m^3/kg
Using the ideal gas law, we can find the density of ice at the final pressure and specific volume:
ρ_f = p_f / (R * T_f * v_f)
where R is the gas constant for water vapor (461.5 J/kg-K).
Assuming that the pressure remains close to atmospheric pressure during the phase change, we can use the initial pressure and temperature to estimate the final pressure and temperature:
T_f ≈ 273 K
p_f ≈ 101.3 kPa + 15.8 MPa/K * (273 K - 273 K) = 101.3 kPa
ρ_f = 101.3 kPa / (461.5 J/kg-K * 273 K * 0.001091 m^3/kg) ≈ 917 kg/m^3
This is consistent with the density of ice at atmospheric pressure (about 917 kg/m^3).
dp/dT = ΔH / TΔV
where dp/dT is the rate of change of pressure with respect to temperature, ΔH is the latent heat of the phase change, T is the temperature, and ΔV is the change in specific volume.
At the freezing point of water (273 K), we can solve for the pressure at which ice would form:
dp/dT = (3.66 * 10^5 J/kg) / (273 K * 91 * 10^-6 m^3/kg) = 15.8 MPa/K
To find the pressure at which ice would form, we need to integrate this equation from the initial pressure (i.e. atmospheric pressure) to the pressure at which ice forms. We can assume that the initial specific volume of water is 0.001 m^3/kg (since water is essentially incompressible at atmospheric pressure).
∫dp = ∫(15.8 MPa/K dT) from 273 K to T_f
p_f - 101.3 kPa = 15.8 MPa/K * (T_f - 273 K)
Solving for the pressure at which ice would form (p_f):
p_f = 101.3 kPa + 15.8 MPa/K * (273 K - T_f)
We can use the fact that the change in specific volume during freezing is 91 x 10^-6 m^3/kg to find the final specific volume of ice:
v_f = v_i + Δv = 0.001 m^3/kg + 91 x 10^-6 m^3/kg = 0.001091 m^3/kg
Using the ideal gas law, we can find the density of ice at the final pressure and specific volume:
ρ_f = p_f / (R * T_f * v_f)
where R is the gas constant for water vapor (461.5 J/kg-K).
Assuming that the pressure remains close to atmospheric pressure during the phase change, we can use the initial pressure and temperature to estimate the final pressure and temperature:
T_f ≈ 273 K
p_f ≈ 101.3 kPa + 15.8 MPa/K * (273 K - 273 K) = 101.3 kPa
ρ_f = 101.3 kPa / (461.5 J/kg-K * 273 K * 0.001091 m^3/kg) ≈ 917 kg/m^3
This is consistent with the density of ice at atmospheric pressure (about 917 kg/m^3).
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The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. Can you explain this answer?
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The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. 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 The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. Can you explain this answer?.
The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. 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 The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The change in specific volume when 1 kg of water freezes is 91 x 10-6 m3. The pressure at 273 K i.e. Find out the pressure at which the ice would freeze, when it is given that latent heat of ice = 3.66 * 105 J/kg & one atmosphere=10-5Correct answer is '136.2'. Can you explain this answer?.
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