Describe the relationship between temperature and specific volume?

In summary: My second...In summary, the conversation revolves around an experiment involving heating a fixed quantity of water in a cylinder fitted with a movable piston. The temperature and volume of the system increase, but the volume does not significantly increase until gas bubbles begin to form. The volume increases at a slower rate when compared to heating liquid water. The experiment is then repeated at different pressures, and the pressure is kept constant. Questions are raised about the properties of water and steam and the experimental setup. The expert is asked about the trends in the Temperature vs specific volume plot for water at constant and different pressures and whether the curves will be less or more steep when the experiment is carried out at a higher external pressure.
  • #1
AAMAIK
47
0
TL;DR Summary
I am trying to understand the trends in the Temperature vs specific volume plot for water at constant and different pressures
I set up an experiment where I put a fixed quantity of water in a cylinder fitted with a movable piston. I slowly add heat to the system, as expected both the temperature and the volume will increase but the volume will not increase significantly (steep line), until some point where some gas bubbles begin to form, Volume increases most shallow (Temperature=constant) if I plot a graph of Temperature vs Volume. Heating beyond the point(last drop of liquid)results in a less steep curve in comparison to heating liquid water at the point where the first bubble vapor forms I add equal amounts of heat to a vapor and a liquid of the same pure substance keeping the pressure=constant the change in the gaseous state will be higher, vapor molecules have more kinetic energy thus expand at a higher rate. But repeating the same experiment at different pressures at the same initial temperature as the one conducted previously. In one case the pressure above the piston is higher than atmospheric pressure then the volume at the initial state is slightly lower. Qualitatively the same thing will happen but does vaporization occcur at a higher temperature and higher specific volume because the pressure the vapor molecules push against is higher so they must have enough K.E to do and once these molecules acquire such K.E they must have a higher velocity than at the saturation temperature for the experiment conducted previously? If I do a force balance on the piston how is the pressure =constant although more vapor molecules are produced and thus pressure must increase?
 
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  • #2
Wow, that paragraph is very difficult to follow. I'm not sure if you are asking about the properties of water and steam. The properties are what they are, you look them up in a table.

Your description also gives me a lot of doubt about the experimental methods. For example, how do you know you add "equal amounts of heat"? Is there air, or steam in the piston, or strictly liguid? Some graphics to explain your experimental setup would be helpful.

But I will ping our thermo expert @Chestermiller to see if he can help.
 
  • #3
anorlunda said:
Wow, that paragraph is very difficult to follow. I'm not sure if you are asking about the properties of water and steam. The properties are what they are, you look them up in a table.

Your description also gives me a lot of doubt about the experimental methods. For example, how do you know you add "equal amounts of heat"? Is there air, or steam in the piston, or strictly liguid? Some graphics to explain your experimental setup would be helpful.

But I will ping our thermo expert @Chestermiller to see if he can help.
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  • #4
AAMAIK said:
Summary: I am trying to understand the trends in the Temperature vs specific volume plot for water at constant and different pressures
But repeating the same experiment at different pressures at the same initial temperature as the one conducted previously. In one case the pressure above the piston is higher than atmospheric pressure then the volume at the initial state is slightly lower. Qualitatively the same thing will happen but does vaporization occcur at a higher temperature and higher specific volume because the pressure the vapor molecules push against is higher so they must have enough K.E to do and once these molecules acquire such K.E they must have a higher velocity than at the saturation temperature for the experiment conducted previously? If I do a force balance on the piston how is the pressure =constant although more vapor molecules are produced and thus pressure must increase?
I don't understand what you are asking. I don't see what you are so puzzled about with regard to this experiment. You refer to a force balance on the piston. Maybe you can elaborate on your doubt?
 
  • #5
Chestermiller said:
I don't understand what you are asking. I don't see what you are so puzzled about with regard to this experiment. You refer to a force balance on the piston. Maybe you can elaborate on your doubt?
My first doubt is whether the curve joining the point (saturated vapor)and the points in the superheated region is less steep than the curve joining the points in the compressed liquid region and the point(saturated liquid). Does the greater kinetic energy of vapor to expand explain why this is the case?My second doubt If we carry out the experiment at a higher external pressure starting at the same initial temperature and slowly add heat.Qualitatively the curves will look similar but will the curves be less/more steep than the curves generated from the same experiment but carried out at a lower temperature and why?
 
  • #6
AAMAIK said:
My first doubt is whether the curve joining the point (saturated vapor)and the points in the superheated region is less steep than the curve joining the points in the compressed liquid region and the point(saturated liquid). Does the greater kinetic energy of vapor to expand explain why this is the case?
A liquid is held together with attractive forces between the molecules which prevent the volume of a liquid to grow more than just a little with changes in pressure and temperature. The molecules of a vapor are far enough apart that these forces are negligible on average (i.e., except for vapor molecules that come very close to one another during their travels).

My second doubtIf we carry out the experiment at a higher external pressure starting at the same initial temperature and slowly add heat.Qualitatively the curves will look similar but will the curves be less/more steep than the curves generated from the same experiment but carried out at a lower temperature and why?
I think in your last sentence you meant "lower pressure."
Can you please provide a sketch of what you mean?
 
  • #7
Chestermiller said:
I think in your last sentence you meant "lower pressure."
Can you please provide a sketch of what you mean?
I am sorry, yes I mean lower pressure. Will the constant pressure line in the T-V plot below for a higher pressure be less/more steep than the constant lower pressure line. If it is so is it for the reason I mentioned in my first doubt, that is at higher pressure molecules must gain enough kinetic energy to turn into vapor and push against the external pressure imposed upon thus vaporization at a higher external pressure will take place at a higher temperature and higher specific volume. Secondly, the gap between the point of saturated liquid and saturated vapor narrows down as the external pressure goes up . External pressure is a parameter that I can control so if I increase the external pressure my system is a compressed liquid and I start to slowly add heat to my system, liquid molecules gain enough energy to lift the piston, but for a certain high external pressure molecules would have to posses high energy to overcome and lift the piston and that energy is high enough that change in state from liquid to vapor occurs a small range of a small specific volume. I am sorry If I can't articulate myself clearly.
 

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  • #8
My third doubt, is why is the volume changing in my T-V plot if I am starting with compressed liquid and I am imposing my system to a certain external pressure. To lift the piston the vapor formed from slowly adding heat must push against the external pressure and once the pressure caused by the vapor molecules must equate or be greater than the external pressure. Does the pressure caused by the vapor molecules equal the external pressure imposed upon the system or is it higher because as I slowly add heat more vapor molecules there are more vapor molecules (in number and energy) hitting the bottom layer of the piston. If its equal to the external pressure all along the process because the volume is changing and thus the gain in kinetic energy compensates for the higher distance molecules would have to travel to hit the piston. I wrote a force balance the pressure caused the vapor molecules must equal the external pressure all along the process if it's in quasi-equilibrium but I am trying to reason why
 
  • #9
Well, I think the first doubt is related to the ideal gas law. It tells us that $$PV=nRT$$where P is the pressure, V is the vapor volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Are you familiar with this law? If we rearrange this equation a little, we get $$v=\frac{V}{n}=\frac{RT}{P}$$where v is the specific (molar) volume. You can see from this equation that, at a constant pressure, the slope of v vs T decreases as the constant pressure gets higher. So the real question is , "why does a vapor obey the ideal gas law? or what is the derivation of the ideal gas law." You can answer this by Googling "derivation of the ideal gas law."

With regard to your third question, if, at any point, you stopped adding heat, the piston would be in equilibrium. Adding heat very slowly does not significantly alter this situation. So, for a massless frictionless piston, the gas pressure will always be essentially equal to the external pressure. So the increased kinetic energy of the molecules is offset by the greater distances between molecules (because of the increased volume) which results in a reduced frequency of collisions with the piston.

Your 2nd question related to the effect of temperature on the behavior of the liquid and vapor. As the temperature increases (and approaches the critical temperature), the liquid behaves more and more like a vapor, and its specific volume increases substantially (as the molecules get further apart). At the critical temperature, the liquid and vapor are indistinguishable, and there is no change in specific volume. Beyond the critical temperature, there is only one phase present (irrespective of the pressure), called the supercritical phase.
 
  • #10
Chestermiller said:
With regard to your third question, if, at any point, you stopped adding heat, the piston would be in equilibrium. Adding heat very slowly does not significantly alter this situation. So, for a massless frictionless piston, the gas pressure will always be essentially equal to the external pressure. So the increased kinetic energy of the molecules is offset by the greater distances between molecules (because of the increased volume) which results in a reduced frequency of collisions with the piston.

If I stopped adding heat, the piston will not move instead it will remain still.

Chestermiller said:
Your 2nd question related to the effect of temperature on the behavior of the liquid and vapor. As the temperature increases (and approaches the critical temperature), the liquid behaves more and more like a vapor, and its specific volume increases substantially (as the molecules get further apart). At the critical temperature, the liquid and vapor are indistinguishable, and there is no change in specific volume. Beyond the critical temperature, there is only one phase present (irrespective of the pressure), called the supercritical phase.
So do you mean that at the critical temperature all the bonds between the liquid molecules break, not over a certain specific volume range but instead vaporization occurs instantaneously described by a point called the critical point? How can I describe the critical point? From what I understood of your reply is that no two phases exist, so is all the system composed of a phase that alternates between all vapor to all liquid?
 
  • #11
AAMAIK said:
If I stopped adding heat, the piston will not move instead it will remain still.
Yes. So?

So do you mean that at the critical temperature all the bonds between the liquid molecules break, not over a certain specific volume range but instead vaporization occurs instantaneously described by a point called the critical point? How can I describe the critical point? From what I understood of your reply is that no two phases exist, so is all the system composed of a phase that alternates between all vapor to all liquid?
At the critical temperature and beyond, only a single phase can exist. It does not alternate between liquid and vapor. it just can't be distinguished as being one or the other, and is merely called a gas. If you want to say that this has more vapor character than liquid character, that's fine.
 
  • #12
Chestermiller said:
yes. So?
Nothing, its clear,I just wanted to make sure I understood what happened if I stopped adding heat slowly.

Chestermiller said:
At the critical temperature and beyond, only a single phase can exist. It does not alternate between liquid and vapor. it just can't be distinguished as being one or the other, and is merely called a gas. If you want to say that this has more vapor character than liquid character, that's fine.
But at lower temperatures, the system is part vapor and part liquid. I can't get the description of the state at the critical point, namely that vapor and liquid can't be distinguished. It's either all liquid, all vapor, or part liquid part vapor.
Thank you for answering and clearing my doubts :)
 

1. How does temperature affect specific volume?

As temperature increases, the specific volume of a substance also increases. This means that the substance takes up more space for the same amount of mass. This relationship is known as thermal expansion.

2. What is the mathematical relationship between temperature and specific volume?

The mathematical relationship between temperature and specific volume is described by the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. This equation shows that as temperature increases, volume also increases for a fixed amount of gas.

3. How is specific volume affected by changes in temperature?

Changes in temperature can have a significant impact on specific volume. As temperature increases, the molecules of a substance gain more kinetic energy and move faster, causing them to take up more space. This results in an increase in specific volume.

4. Is there a limit to how much specific volume can change with temperature?

Yes, there is a limit to how much specific volume can change with temperature. This limit is known as the critical temperature, which is the temperature at which a substance can no longer exist as a liquid and becomes a gas. Above this temperature, further increases in temperature will not lead to a significant increase in specific volume.

5. How does the relationship between temperature and specific volume differ for different substances?

The relationship between temperature and specific volume can vary for different substances. In general, most substances follow the ideal gas law, but some substances, such as liquids and solids, may exhibit different behavior due to intermolecular forces. For example, water has a unique relationship between temperature and specific volume, with a maximum density at 4°C before decreasing as temperature increases.

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