Ideal Gas/Work Done by Water Problemq

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Homework Help Overview

The problem involves calculating the work done by water in a sealed container as it is heated from room temperature to a higher temperature. The context is centered around the behavior of water transitioning from a liquid to a gas, with specific parameters provided such as mass, molar mass, and density.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law to determine the volume of water, questioning the appropriateness of using gas equations for a liquid. There are inquiries about the correct equations to use and the relationship between pressure, volume change, and work done.

Discussion Status

The discussion is ongoing, with participants providing feedback on the initial attempts and suggesting alternative approaches. Some participants are exploring the implications of using the gas law for the final state of water and the density for the initial state.

Contextual Notes

There is a noted confusion regarding the transition from liquid to gas and the assumptions made about pressure and volume changes. The problem constraints include the need to consider the external air pressure and the specific conditions of the water's state during heating.

Bgerst103
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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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