Ideal Gas/Work Done by Water Problemq

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In summary, using the gas law for the final state and the density of water for the initial state, the work done by the water molecules as they boil from room temperature to 200 degrees C is equal to 161000 kJ. This is the amount of work done because it is the product of the outside air pressure and the change in volume of the water, which is the amount of force the water exerts on its surroundings.
  • #1
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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  • #2
First thing you did wrong was to use the GAS law to find the volume of LIQUID water.
 
  • #3
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?
 
  • #4
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
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.
 

1. What is an ideal gas?

An ideal gas is a theoretical gas composed of particles that have no volume and do not interact with each other. It follows the ideal gas law, which describes the relationship between pressure, volume, temperature, and number of moles of an ideal gas.

2. How is work done by water calculated in a problem?

In a problem involving work done by water, the formula used is W = PΔV, where W represents work, P is the pressure of the water, and ΔV is the change in volume of the system.

3. What is the difference between an ideal gas and a real gas?

An ideal gas is a theoretical concept, while a real gas is an actual gas with particles that have volume and interact with each other. Real gases do not always follow the ideal gas law, especially at high pressures and low temperatures.

4. How does the ideal gas law relate to the work done by water?

The ideal gas law, PV = nRT, can be used to calculate the pressure and volume of an ideal gas in a problem involving work done by water. This information is then used in the formula W = PΔV to determine the work done by the water.

5. Can the ideal gas law be used for any gas?

No, the ideal gas law is only applicable to ideal gases, which are theoretical gases composed of non-interacting particles. Real gases may deviate from the ideal gas law, especially at high pressures and low temperatures.

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