The heat of vaporization of water is 9.73
kcal/mol. At what pressure (in torr) would
pure water boil at 89◦C?
Use the Clausius-Clapeyron equation. You can use the "other" p/t numbers as 273 K at 760 torr. Remember to use the correct value for R.
To determine the pressure at which water boils at a specific temperature, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
- P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
- ΔHvap is the heat of vaporization.
- R is the ideal gas constant (8.314 J/(mol·K)).
- T1 and T2 are the temperatures in Kelvin.
First, we need to convert the given temperature of 89◦C to Kelvin:
T2 = 89 + 273.15 = 362.15 K
Next, we can plug the values into the formula and solve for ln(P2/P1):
ln(P2/P1) = (-9.73 kcal/mol / (8.314 J/(mol·K))) * (1/362.15 K - 1/373.15 K)
Now we can solve for ln(P2/P1) by simplifying the equation:
ln(P2/P1) = (-0.0221) * (-0.002929)
ln(P2/P1) = 0.000064755
To get the actual value of P2/P1, we can use the natural logarithm inverse function (e^x):
P2/P1 = e^(0.000064755)
P2/P1 = 1.000065
Finally, to find the pressure in torr, we need to convert P2/P1 back to P2:
P2 = P1 * (P2/P1)
P2 = P1 * 1.000065
Since we want to find the pressure in torr, we can assume atmospheric pressure (1 atm) as P1:
P2 = 1 atm * 1.000065
Now we can convert atm to torr (1 atm = 760 torr):
P2 = 760 torr * 1.000065
Therefore, pure water would boil at approximately 760.049 torr at a temperature of 89◦C.