# The Clausius-Clapeyron relation to study pressure cookers

• gabu
In summary, 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
gabu
Please post this type of questions in HW section using the template.
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.

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.)

gabu
gabu said:
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]

## 1. What is the Clausius-Clapeyron relation and how is it related to pressure cookers?

The Clausius-Clapeyron relation is a mathematical equation that describes the relationship between temperature and vapor pressure for a given substance. In pressure cookers, this relation is used to determine the boiling point of water at different pressures, which is crucial for the cooking process.

## 2. How does the Clausius-Clapeyron relation affect the cooking time in a pressure cooker?

The Clausius-Clapeyron relation states that as pressure increases, the boiling point of water also increases. This means that in a pressure cooker, the higher pressure allows for the water to reach a higher temperature, resulting in faster cooking times compared to traditional cooking methods.

## 3. Can the Clausius-Clapeyron relation be used to determine the safety of a pressure cooker?

No, the Clausius-Clapeyron relation does not directly relate to the safety of a pressure cooker. The safety of a pressure cooker depends on the design and construction of the cooker, as well as proper usage and maintenance.

## 4. Does the Clausius-Clapeyron relation apply to all substances in a pressure cooker or just water?

The Clausius-Clapeyron relation can be applied to any substance in a pressure cooker, as long as the substance has a known vapor pressure-temperature relationship. However, in pressure cooking, water is the most commonly used substance, so the relation is most often used for water.

## 5. How does altitude affect the Clausius-Clapeyron relation in pressure cookers?

The Clausius-Clapeyron relation is affected by altitude because as altitude increases, atmospheric pressure decreases. This means that at higher altitudes, the boiling point of water is lower, and therefore the pressure cooker will need to reach a higher pressure to achieve the same cooking temperature. This is why pressure cooking times may need to be adjusted at higher altitudes.

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