The temperature inside a pressure cooker is 115^∘ C. Calculate the vapor pressure of water insid

The temperature inside a pressure cooker is 115^∘ C....

Key Concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation provides a quantitative relationship between the vapor pressure of a substance and its temperature. It is commonly used to calculate how changes in temperature affect vapor pressure, and vice versa, by incorporating the enthalpy of vaporization. This equation is key in understanding and calculating the conditions required for phase changes.

Phase Equilibrium

Phase equilibrium refers to a state where the liquid and vapor phases of a substance coexist at a constant temperature and pressure, with the rate of evaporation equal to the rate of condensation. This balance is essential for defining the boiling point and analyzing conditions in closed systems, such as a pressure cooker.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature. As temperature increases, more molecules escape from the liquid phase into the vapor phase, raising the vapor pressure. Understanding vapor pressure is crucial in studying how liquids evaporate and boil under different conditions.

Transcript

00:01 We are given information about the temperature inside of a pressure cooker, and we want to use the closius clapperon equation in order to help us determine what the vapor pressure of water is inside the pressure cooker.

00:15 So this is the equation that we are given, and we can rearrange this into the more familiar form of the clausius clapperon equation by subtracting over this term to the left side and this term to the right side.

00:29 And then when we factor out appropriately, we get the expression, the natural log of the vapor pressure at t1, divided by the vapor pressure at t2, is equal to delta h of vaporization over r times one over t2 minus one over t1.

01:13 And we know that normally water has a boiling point of 100 degrees celsius, and at its boiling point, its vapor pressure is equal to the atmospheric pressure.

01:24 So that's how we know values for t1 and the vapor pressure at t1.

01:30 And then we are told what the temperature t2 is inside of the pressure cooker.

01:35 So all we have to do is use that equation to solve for the vapor pressure at that temperature of water.

01:41 And we can look up the value for the delta h of vaporization for water, and we know what the constant r is.

01:50 So we rearrange this equation out of solve for pvap of t2.

01:56 So the vapor pressure at t2, which is the vapor pressure of water in the pressure cooker at a temperature of 115 degrees celsius, is equal to the vapor pressure of water at t1, which is its boiling point, divided by e to the power of, when we take e to the power of both sides, we get rid of that ln, delta hvap over r times 1 over t2 minus 1 over t1.

02:45 And now we plug in everything that we know into this equation to solve for the vapor pressure of water inside that pressure cooker.

02:58 So the vapor pressure at t1 is one atmosphere, since that's equal to the atmospheric pressure at the boiling point of water.

03:06 So that goes on the numerator.

03:10 We divide this by e to the power of that delta h of vaporization, which we can convert into joules per mole, so we can cancel off units for the constant r.

03:23 So that is 40 .7 times 10 to the third joules per mole divided by r, which is 8 .314 joules per mole times kelvin.

03:48 We multiply this by 1 over t2, which was given in the problem as 115 degrees celsius, which is the temperature and the pressure cooker...

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