Pressure Cooker-Involving Ideal Gas law and Total Pressure

[tcw46]

Homework Statement

A cook puts 9.00g of water in a 2.00L pressure cooker and warms it to 500 degree celsius. What is the pressure inside the container.

Homework Equations

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

PV = nRT (eqn2)

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

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

 

[nickjer]

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