The Clausius-Clapeyron relation to study pressure cookers

[gabu]

The Clausius-Clapeyron formula is given by

$\frac{d P}{d T} = \frac{L}{T \Delta V}$

where P and T are the pressure and temperature at the boiling point, respectively, and L is the latent heat per mole at the boiling point, and $\Delta V$ is the change in the volume per mole between the gas and liquid phases. For water, find the value of $(dP/dT)_{T=T_{0}}$ when the pressure is 1 atmosphere and the temperature the boiling temperature $T_{0}$. Consider the latent heat to be $L = 540cal/g$ and the gas constant as $R=2.0\,cal\, mol^{-1}K^{-1}$

b) If the temperature of air and water inside the pressure cooker prior to heating is $T_{1}(<T_{0})$ , then how many atmospheres does the internal pressure of the cooker P reach when the temperature due to heating increases to T? Assume the water doesn't boil
Now... for the first part of the question I only have to determine the variation in volume at the boiling point so I can calculate the quantity asked. My idea to determine such variation is to use that

$P\Delta V = R T$

but I'm not sure if I can use this equation for water vapour. Can it be considered an ideal gas? My idea is that, if it can, the volume per mole of vapour under those conditions is the variation in the volume of water.

For the second question, my idea is to integrate Clausius-Clapeyron equation but, I don't know the relation between $\Delta V$ and T or P. The only relation that comes to my mind is the ideal gases relations, but if I am to assume the water does not boil I can't use this relation. Which relation should I use?

Thank you very much.

 

[mjc123]

Water vapour won't be perfectly ideal, but in the absence of any further information, you have to use the ideal gas equation.
In part 2, the water doesn't boil (i.e. boil away), but it does evaporate. There is an equilibrium vapour pressure at every temperature; the water will boil when this pressure equals the external atmospheric pressure, but this can't happen inside a pressure cooker. Again, use the ideal gas equation for the water vapour. (Don't forget to account for the pressure of the air. I think you have to assume the cooker is sealed with P = 1 atm at T₁.)

 

[Chestermiller]

The Clausius-Clapeyron formula is given by

$\frac{d P}{d T} = \frac{L}{T \Delta V}$

where P and T are the pressure and temperature at the boiling point, respectively, and L is the latent heat per mole at the boiling point, and $\Delta V$ is the change in the volume per mole between the gas and liquid phases. For water, find the value of $(dP/dT)_{T=T_{0}}$ when the pressure is 1 atmosphere and the temperature the boiling temperature $T_{0}$. Consider the latent heat to be $L = 540cal/g$ and the gas constant as $R=2.0\,cal\, mol^{-1}K^{-1}$

b) If the temperature of air and water inside the pressure cooker prior to heating is $T_{1}(<T_{0})$ , then how many atmospheres does the internal pressure of the cooker P reach when the temperature due to heating increases to T? Assume the water doesn't boil
Now... for the first part of the question I only have to determine the variation in volume at the boiling point so I can calculate the quantity asked. My idea to determine such variation is to use that

$P\Delta V = R T$

but I'm not sure if I can use this equation for water vapour. Can it be considered an ideal gas?

The Clausius-Clapeyron equation assumes that the vapor behaves as an ideal gas.

My idea is that, if it can, the volume per mole of vapour under those conditions is the variation in the volume of water.

The $\Delta V$ in the equation is the volume of one mole of saturated vapor minus the volume of one mole of saturated liquid. In the derivation of the Clausius-Clapeyron equation, the volume of the saturated liquid is neglected compared to that of the vapor.

For the second question, my idea is to integrate Clausius-Clapeyron equation but, I don't know the relation between $\Delta V$ and T or P. The only relation that comes to my mind is the ideal gases relations, but if I am to assume the water does not boil I can't use this relation.

Who says? All the Clausius-Clapeyron equation gives is the relationship between pressure and temperature for water vapor in thermodynamic equilibrium with saturated liquid vapor. This has nothing directly to do with boiling.[/QUOTE]