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the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure?
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the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liq...
The rate of change of a substance at a specific temperature and pressure can be determined using the Clausius-Clapeyron equation. This equation relates the rate of change of vapor pressure with temperature to the enthalpy of vaporization and the molar volumes of the liquid and vapor phases.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
ΔHvap is the enthalpy of vaporization.
R is the gas constant (8.314 J/mol·K).
T1 and T2 are the initial and final temperatures, respectively.
In this case, we are given the enthalpy of vaporization (ΔHvap) as 40.65 kJ/mol, the molar volume of liquid water as 0.019 × 10^-2 m^3/mol, and the molar volume of steam as 30.199 × 10^-3 m^3/mol. We are asked to calculate the rate of change of water at 100°C and 1 atm.
To apply the Clausius-Clapeyron equation, we need to convert the given volumes to m^3/mol and the enthalpy of vaporization to J/mol.
The molar volume of liquid water is given as 0.019 × 10^-2 m^2/mol. Converting this to m^3/mol:
0.019 × 10^-2 m^2/mol = (0.019 × 10^-2 m^2/mol) * (1 m^3/10^6 cm^3) * (1000 cm^3/1 m^3) = 1.9 × 10^-8 m^3/mol
The molar volume of steam is given as 30.199 × 10^-3 m^3/mol.
The enthalpy of vaporization is given as 40.65 kJ/mol. Converting this to J/mol:
40.65 kJ/mol = 40.65 × 10^3 J/mol
Now, we can plug these values into the Clausius-Clapeyron equation:
ln(P2/1 atm) = (40.65 × 10^3 J/mol / 8.314 J/mol·K) * (1/373 K - 1/373 K)
Simplifying:
ln(P2/1 atm) = (40.65 × 10^3 J/mol / 8.314 J/mol·K) * (0)
Since (0) * anything is zero, the right side of the equation becomes zero. Therefore:
ln(P2/1 atm) = 0
Taking the exponential of both sides:
P2/1 atm = e^0
Since e^0 is equal to 1, we have:
P2/1 atm = 1
Multiplying both sides by 1 atm, we obtain:
P2 = 1 atm
Therefore, the vapor pressure of water at 100°C and 1 atm is 1 atm, and there is no rate of change at that point.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
ΔHvap is the enthalpy of vaporization.
R is the gas constant (8.314 J/mol·K).
T1 and T2 are the initial and final temperatures, respectively.
In this case, we are given the enthalpy of vaporization (ΔHvap) as 40.65 kJ/mol, the molar volume of liquid water as 0.019 × 10^-2 m^3/mol, and the molar volume of steam as 30.199 × 10^-3 m^3/mol. We are asked to calculate the rate of change of water at 100°C and 1 atm.
To apply the Clausius-Clapeyron equation, we need to convert the given volumes to m^3/mol and the enthalpy of vaporization to J/mol.
The molar volume of liquid water is given as 0.019 × 10^-2 m^2/mol. Converting this to m^3/mol:
0.019 × 10^-2 m^2/mol = (0.019 × 10^-2 m^2/mol) * (1 m^3/10^6 cm^3) * (1000 cm^3/1 m^3) = 1.9 × 10^-8 m^3/mol
The molar volume of steam is given as 30.199 × 10^-3 m^3/mol.
The enthalpy of vaporization is given as 40.65 kJ/mol. Converting this to J/mol:
40.65 kJ/mol = 40.65 × 10^3 J/mol
Now, we can plug these values into the Clausius-Clapeyron equation:
ln(P2/1 atm) = (40.65 × 10^3 J/mol / 8.314 J/mol·K) * (1/373 K - 1/373 K)
Simplifying:
ln(P2/1 atm) = (40.65 × 10^3 J/mol / 8.314 J/mol·K) * (0)
Since (0) * anything is zero, the right side of the equation becomes zero. Therefore:
ln(P2/1 atm) = 0
Taking the exponential of both sides:
P2/1 atm = e^0
Since e^0 is equal to 1, we have:
P2/1 atm = 1
Multiplying both sides by 1 atm, we obtain:
P2 = 1 atm
Therefore, the vapor pressure of water at 100°C and 1 atm is 1 atm, and there is no rate of change at that point.
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the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry?
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the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry?.
the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry?.
Solutions for the enthalpy of vapourisationof h2ois 40.65 ki/mol,molar volume of liquid water is 0.019*10^-2 m^2.mol^-1 and molar volume of steam is 30.199*10^-3m^3.mol^-1 all at 100°c and 1 atm.calculate the rate of change in the following point of water at100°c with atmosphere pressure? Related: The Clausius Clapeyron Equation - Thermochemistry? in English & in Hindi are available as part of our courses for Chemistry.
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