# Ideal Gas/Work Done by Water Problemq

1. Apr 27, 2014

### Bgerst103

1. The problem statement, all variables and given/known data

You have a sealed container with 2 kg of water (molar mass: 18 g/mol, density: 1000 kg/m 3). The lid on the the container is very light, and the container side walls are very tall, like a tall graduated cylinder. You start with the water at room temperature 25 degrees C and boil all the water until you get to 200 degrees C. What is the Work done by the water molecules?
Select one:
a. 234 kJ
b. 334 kJ
c. 434 kJ
d. 534 kJ

2. Relevant equations

PV=nRT
P1V1/T1=P1V2/T2
p(deltaV)=nR(deltaT)
-deltaW=deltaE

3. The attempt at a solution

I tried solving for p(deltaV) which is equal to the change in energy which could then be used to find work done but never could get one of those answers. I did this by starting with PV=nRT.
P=100,000 Pa
V=?
n=111.111
R=8.314
T= 25+273.15=298.15
Plug those in to get V=2.75
Plug that into P1V1/T1=P2V2/T2 to get V2=4.36
So then P(deltaV) should equal 100,000 (4.36-2.75)=161000 which equal -deltaW. I don't really know where to go from here or what I'm doing wrong.

2. Apr 27, 2014

### dauto

First thing you did wrong was to use the GAS law to find the volume of LIQUID water.

3. Apr 27, 2014

### Bgerst103

Care to give me the right equation to use then? I saw moles and assumed pv=nrt. Is it mass over density?

4. Apr 27, 2014

### Staff: Mentor

What is the pressure that the outside air is pressing down with on the top of the lid?
What is the change in volume of the water in the container if its pressure always matches the air pressure on the top side of the lid?

What is the product of the outside air pressure and the volume change of the water in the container (in units of kJ)? Why is this the amount of work that the water in the container does?

Chet

5. Apr 27, 2014

### dauto

The final state is a gas but the initial state is a liquid. Use the gas law for the final state and the water density (also provided in the problem) for the initial state.