Knowing that delta Hvap for water is 40.7 kJ/mol, calculate Pvap of water at 46 C
Use the Clausius-Clapeyron equation. Data not listed that you can use is PH2O is 760 mm at 100 C.
Thanks
To calculate the vapor pressure (Pvap) of water at a given temperature, you can use the Clausius-Clapeyron equation:
ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)
where:
- P1 is the vapor pressure at temperature T1
- P2 is the vapor pressure at temperature T2
- ∆Hvap is the enthalpy of vaporization
- R is the gas constant (8.314 J/(mol·K))
- T1 and T2 are the temperatures in Kelvin
First, we need to convert the given temperature of 46 °C to Kelvin:
T2 = 46 + 273.15 = 319.15 K
Assuming we know the vapor pressure of water at a reference temperature (T1), let's say at its boiling point (100 °C):
T1 = 100 + 273.15 = 373.15 K
Now we can substitute these values into the equation and solve for Pvap:
ln(P2/P1) = (40.7 kJ/mol / (1000 J/kJ)) / (8.314 J/(mol·K)) * (1/373.15 K - 1/319.15 K)
Note: We divided ∆Hvap by 1000 to convert it from kJ/mol to J/mol.
Let's calculate the right side of the equation first:
(40.7 kJ/mol / (1000 J/kJ)) / (8.314 J/(mol·K)) * (1/373.15 K - 1/319.15 K) ≈ 0.0548
Now let's solve for ln(P2/P1):
ln(P2/P1) = 0.0548
To find Pvap, we need to exponentiate to isolate P2/P1:
P2/P1 = e^(0.0548)
Finally, we can calculate Pvap by multiplying P1 (vapor pressure at the reference temperature) by P2/P1:
Pvap = P1 * e^(0.0548)
The value of P1 depends on the reference source used, such as a reference book or scientific source. Please provide the value of P1, and I can help you calculate the vapor pressure at 46 °C.
To calculate the vapor pressure (Pvap) of water at 46°C, we need to use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = heat of vaporization
R = gas constant (8.314 J/(mol·K))
T1 = initial temperature (in Kelvin)
T2 = final temperature (in Kelvin)
First, let's convert the given temperature of 46°C to Kelvin:
T2 = 46°C + 273.15 = 319.15 K
Now we can substitute the values into the equation:
ln(P2/P1) = (40.7 kJ/mol / 8.314 J/(mol·K)) * (1/373.15 K - 1/319.15 K)
Simplifying further:
ln(P2/P1) = 4.903 * (0.002681 - 0.003130)
ln(P2/P1) = 4.903 * (-0.00045)
ln(P2/P1) ≈ -0.002206
Next, we need to rearrange the equation to solve for P2/P1:
P2/P1 = e^(ln(P2/P1))
P2/P1 ≈ e^(-0.002206)
P2/P1 ≈ 0.997796
Finally, we multiply both sides of the equation by P1 to isolate P2:
P2 ≈ 0.997796 * P1
Now, to get the value of P1, we can look up the vapor pressure of water at 100°C, which is a commonly known value:
P1 (vapor pressure of water at 100°C) ≈ 101.325 kPa (or 1 atm)
Substituting P1 and solving for P2:
P2 ≈ 0.997796 * 101.325 kPa
P2 ≈ 100.947 kPa (rounded to three decimal places)
Therefore, the estimated vapor pressure of water at 46°C is approximately 100.947 kPa.