Is it possible to calculate heat of vaporization if i only have pressure and temperature.
It is somewhat complicated, but there is good theory.
http://pubs.acs.org/doi/abs/10.1021/ja01212a080
Yes, it is possible to calculate the heat of vaporization if you have the pressure and temperature. The heat of vaporization, also known as the enthalpy of vaporization, is the amount of heat energy required to convert a substance from its liquid state to its gaseous state at a specific temperature and pressure.
To calculate the heat of vaporization, you can use the Clausius-Clapeyron equation, which relates the heat of vaporization to the pressure and temperature. The equation is as follows:
ln(P2/P1) = (-ΔHvap/R)*((1/T2) - (1/T1))
In this equation:
- P1 and T1 are the pressure and temperature in the liquid state.
- P2 and T2 are the pressure and temperature in the gaseous state.
- ΔHvap is the heat of vaporization.
- R is the gas constant.
By rearranging the equation, you can solve for the heat of vaporization (ΔHvap):
ΔHvap = -R*((1/T2) - (1/T1))/(ln(P2/P1))
To use this equation, you need to know the values of P1, T1, P2, and T2. Make sure that the temperatures are in Kelvin (K) and the pressures are in the same units (e.g., Pa, atm). The gas constant (R) is a known value and is approximately 8.314 J/(mol·K).
Once you have the values for P1, T1, P2, and T2, you can plug them into the equation to calculate the heat of vaporization (ΔHvap).