Ideal Gas/Work Done by Water Problemq

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The discussion revolves around calculating the work done by water in a sealed container as it transitions from liquid to gas when heated from 25°C to 200°C. The initial approach incorrectly applied the ideal gas law to liquid water, leading to confusion about the correct equations to use. Participants emphasize using the gas law for the final gaseous state and the density of water for the initial liquid state. Key questions include determining the change in volume and the external pressure acting on the container's lid. The correct method involves calculating the work based on the pressure of the outside air and the volume change of the water.
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Homework Statement



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

Homework Equations



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

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

Care to give me the right equation to use then? I saw moles and assumed pv=nrt. Is it mass over density?
 
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
 
Bgerst103 said:
Care to give me the right equation to use then? I saw moles and assumed pv=nrt. Is it mass over density?

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.
 
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