The normal boiling point of water is 100.00 degrees celsius and the enthalpy of vaporization at this temperature is 40.657kj/mol.
What would be the boiling point of water if it were based on a pressure of 1 bar instead of the standard atm?
nvm i figured it out
To determine the boiling point of water at a pressure of 1 bar instead of the standard atmosphere (1 atm), we need to use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at two different temperatures and can be expressed as follows:
ln(P1/P2) = -(ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = enthalpy of vaporization
R = ideal gas constant (8.314 J/mol·K)
T1 = initial temperature (boiling point at 1 atm)
T2 = desired temperature (boiling point at 1 bar)
In this case, we know the boiling point of water at 1 atm is 100.00 degrees Celsius (373.15 Kelvin) and the enthalpy of vaporization is 40.657 kJ/mol. Since 1 bar is equivalent to 0.9869 atm, we need to find the boiling point at this pressure.
Let's substitute the given values into the equation and solve for T2:
ln(1/0.9869) = -(40.657 kJ/mol / (8.314 J/mol·K)) * (1/373.15 K - 1/T2)
Simplifying this equation, we get:
ln(1.0133) = -(4.9004 / (373.15 - T2))
Now, we need to solve for T2. To do that, we isolate the temperature variable:
1.0133 = exp(-(4.9004 / (373.15 - T2)))
Next, we take the natural logarithm of both sides to get rid of the exponential term:
ln(1.0133) = -(4.9004 / (373.15 - T2))
Finally, we solve for T2:
T2 = 373.15 - (4.9004 / ln(1.0133))
Using a calculator, we find that T2 ≈ 99.63 degrees Celsius.
Therefore, the boiling point of water at a pressure of 1 bar is approximately 99.63 degrees Celsius.