# Pressure Cooker-Involving Ideal Gas law and Total Pressure

1. May 10, 2010

### tcw46

1. The problem statement, all variables and given/known data
A cook puts 9.00g of water in a 2.00L pressure cooker and warms it to 500 degree celcius. What is the pressure inside the container.

2. Relevant equations
$$P_{total} = P_{1} + P_{2} +...$$ (eqn1)

$$PV = nRT$$ (eqn2)

$$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{1}}$$ (eqn3)

3. The attempt at a solution
I've assumed that the initial pressure and temperature are standard conditions (101 kPa & 298k)

Using the Total Pressure law, I know that $$P_{i Total} = P_{atm/air} \; and \;P_{f Total} = P_{air,\, heated\, to\, 773K} + P_{Water}$$

$$Total Volume = 2.00L = 0.00200m^3$$

$$T_{initial} = 298K \; and \; T_{final} = 773K$$

using eqn3 i found $$P_{air} = 261990Pa$$ by substituting relative values. (using P1 = 101x1^3 Pa)

at the beginning the pressure of water can be neglected, since it's a liquid.
However at 773K, i use PV=nRT to calculate the partial pressure of water.

n=9.00g/18gmol^-1 = 0.5mol T = 773K V = 0.002m^3 and R = 8.314 J mol^-1 K^-1
Partial Pressure of Water = 1606680 Pa.

Therefore total pressure = sum of the two partial pressure = 1868670 Pa or 1869 kPa.

I'm just wondering whether someone could check this over for me and give me any comments, since i'm a bit nervous about the huge answer :S

Thanks

2. May 11, 2010

### nickjer

Seems ok to me. Unless I made the same mistake you did :)