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Spherical bubbles of 3 mm diameter are observed in the bulk fluid boiling of water at standard atmospheric pressure. Assuming pure water vapor in the bubble and vapor pressure equal to 101.325 k N/m2, calculate the temperature of the vapor
- a)100.217 degree Celsius
- b)200.217 degree Celsius
- c)300.217 degree Celsius
- d)400.217 degree Celsius
Correct answer is option 'A'. Can you explain this answer?
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Spherical bubbles of 3 mm diameter are observed in the bulk fluid boil...
T v – T sat = (2 σ/r – p g) R v T V2/p v h f g.
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Spherical bubbles of 3 mm diameter are observed in the bulk fluid boil...
Calculation of Vaporization Temperature of Water
Given:
Diameter of bubble (d) = 3 mm = 0.003 m
Vapor Pressure of water (Pv) = 101.325 kN/m2
Formula used:
We will use the Laplace's equation to calculate the vaporization temperature of water.
Laplace's equation:
Pv = 2σ/r
Where,
Pv = Vapor Pressure
σ = Surface Tension
r = radius of the bubble
Surface Tension of water (σ) at room temperature is 71.97 mN/m.
Calculation:
We can calculate the radius of bubble (r) from its diameter (d) as follows:
r = d/2 = 0.003/2 = 0.0015 m
Substituting the given values in Laplace's equation:
101.325 × 10^3 = 2 × 71.97 × 10^-3 / 0.0015
101.325 × 10^3 = 95.96 × 10^3 / r
r = 95.96 × 10^3 / 101.325 × 10^3
r = 0.946 m
We can now use the Clausius-Clapeyron equation to calculate the vaporization temperature of water:
ln(Pv/P) = ΔHvap/R(1/Tvap-1/T)
Where,
Pv = Vapor Pressure
P = Atmospheric Pressure
ΔHvap = Enthalpy of vaporization
R = Gas Constant
Tvap = Vaporization Temperature
T = Temperature of water
Assuming the enthalpy of vaporization of water (ΔHvap) at 100 °C to be 40.67 kJ/mol and the gas constant (R) to be 8.314 J/mol-K, we can rewrite the equation as:
ln(101.325 × 10^3/101.325 × 10^3) = 40.67 × 10^3 / 8.314 (1/Tvap - 1/373.15)
Simplifying the equation, we get:
1/Tvap = 1/373.15 + ln(101.325 × 10^3/101.325 × 10^3) × 8.314 / 40.67 × 10^3
1/Tvap = 1/373.15
Tvap = 373.15 K - 273.15
Tvap = 100.217 °C
Therefore, the temperature of the vapor in the bubble is 100.217 °C.
Given:
Diameter of bubble (d) = 3 mm = 0.003 m
Vapor Pressure of water (Pv) = 101.325 kN/m2
Formula used:
We will use the Laplace's equation to calculate the vaporization temperature of water.
Laplace's equation:
Pv = 2σ/r
Where,
Pv = Vapor Pressure
σ = Surface Tension
r = radius of the bubble
Surface Tension of water (σ) at room temperature is 71.97 mN/m.
Calculation:
We can calculate the radius of bubble (r) from its diameter (d) as follows:
r = d/2 = 0.003/2 = 0.0015 m
Substituting the given values in Laplace's equation:
101.325 × 10^3 = 2 × 71.97 × 10^-3 / 0.0015
101.325 × 10^3 = 95.96 × 10^3 / r
r = 95.96 × 10^3 / 101.325 × 10^3
r = 0.946 m
We can now use the Clausius-Clapeyron equation to calculate the vaporization temperature of water:
ln(Pv/P) = ΔHvap/R(1/Tvap-1/T)
Where,
Pv = Vapor Pressure
P = Atmospheric Pressure
ΔHvap = Enthalpy of vaporization
R = Gas Constant
Tvap = Vaporization Temperature
T = Temperature of water
Assuming the enthalpy of vaporization of water (ΔHvap) at 100 °C to be 40.67 kJ/mol and the gas constant (R) to be 8.314 J/mol-K, we can rewrite the equation as:
ln(101.325 × 10^3/101.325 × 10^3) = 40.67 × 10^3 / 8.314 (1/Tvap - 1/373.15)
Simplifying the equation, we get:
1/Tvap = 1/373.15 + ln(101.325 × 10^3/101.325 × 10^3) × 8.314 / 40.67 × 10^3
1/Tvap = 1/373.15
Tvap = 373.15 K - 273.15
Tvap = 100.217 °C
Therefore, the temperature of the vapor in the bubble is 100.217 °C.
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Spherical bubbles of 3 mm diameter are observed in the bulk fluid boiling of water at standard atmospheric pressure. Assuming pure water vapor in the bubble and vapor pressure equal to 101.325 k N/m2, calculate the temperature of the vapora)100.217 degree Celsiusb)200.217 degree Celsiusc)300.217 degree Celsiusd)400.217 degree CelsiusCorrect answer is option 'A'. Can you explain this answer?
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