Thermodynamics: A process that causes P&V to both increase

In summary: Assuming then the gas is in a cylinder that somehow compensates for ΔPΔV for the change in T, ΔT. However, I believe that there exist many combinations such that (ΔPΔV)/Nk = ΔT.Hence the equation requires some form of regulating constant such that ΔP:ΔV (in atm and L)= 1:1 (or equivalent constant ratios for that matter)...The solution is to have a second piston, placed at a distance r2 from the first, and have a spring constant k such thatkΔx/A= ΔP
  • #1
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Homework Statement


Given that P & V increase in direct proportion such that
When P= 1 atm, V= 1L; P=3 atm, V=3L

Describe a possible setup for the pressure to rise as the volume increases.

Homework Equations


F/A=P
V= 2pi.r.h

The Attempt at a Solution


Suppose a system of
1. a cylinder with volume V, height x and radius r
2. with piston with area A and radius r on top
3. and spring with spring constant k on top the piston

Since pressure acts similarly in all directions, assume that the pressure P acts on A such that
(-kx)/A = P

Hence, if

ΔV= 2pi.r.Δx → Δx= ΔV/(2pi.r) ---->(1)

and

ΔF/A= ΔP
kΔx/A= ΔP, where A= pi.r2 ---->(2)

Therefore substitute (1) into (2) for Δx, spring constant k is selected such that it fulfills

k = 2pi2.r3. ΔP/ΔV

The basic idea is that Fspring/A = P and at the same time it also fulfills -1/2kx2= Wvolume,pressure for when both equations uses the same P and V.

I have thought of the compression of air but chose to make it negligible. The idea led from me considering that an increase in pressure when volume is increasing must come from a form of suppression/resistance.

However, it is also to be noted that the main source of energy input would be from heat energy, Q.

Would this make sense? Thanks for reviewing.
 
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  • #2
Hi CoinToss, welcome to PF.

The problem meant P as the pressure of some gas and V its volume. Yes, it can be imagined that the gas is confined in a cylinder, equipped with a movable piston, and the force of the gas from inside is balanced by an equal force from outside, with a spring, for example.

The main problem is what makes the gas expand and its pressure increase at the same time? You can assume that it is an ideal gas. What are the variables characterising the state of the gas in addition to its volume and pressure?

ehild
 
  • #3
Hello ehild! Thanks for the welcome and reviewing the items above.

In addition to volume and pressure, temperature T is in proportion to PV as shown by PV=NkT. T could be proportional to either one or both of the factors depending on whether the other is a constant or no.

Assuming then the gas is in a cylinder that somehow compensates for ΔPΔV for the change in T, ΔT. However, I believe that there exist many combinations such that (ΔPΔV)/Nk = ΔT.
Hence the equation requires some form of regulating constant such that ΔP:ΔV (in atm and L)= 1:1 (or equivalent constant ratios for that matter)...

Am I following your lead correctly?

Important note: I reviewed the question again and a condition provided was that the gas is confined in a cylinder. Sorry to have left that out.
 
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  • #4
CoinToss said:
In addition to volume and pressure, temperature T is in proportion to PV as shown by PV=NkT. T could be proportional to either one or both of the factors depending on whether the other is a constant or no.

Assuming then the gas is in a cylinder that somehow compensates for ΔPΔV for the change in T, ΔT. However, I believe that there exist many combinations such that (ΔPΔV)/Nk = ΔT.
Hence the equation requires some form of regulating constant such that ΔP:ΔV (in atm and L)= 1:1 (or equivalent constant ratios for that matter)...

Am I following your lead correctly?

Yes, so figure out how the temperature should change so as both V and P stay proportional to each other. Imagine that you have 1 mol of ideal gas.
What is the temperature if P=1 atm, V=1 L? And what should be the temperature if P=3 atm, V=3 L?

ehild
 
  • #5
The answers would be T= 12.18K and T= 109.62K respectively.

... However, I apologize for not seeing the relevance from here on out, as I am at a loss for another way such that ΔT directly dictates the ratio of P:V=1:1.
I am thinking that for any value of T for PV=nRT, P:V is not necessarily 1:1. Please advice, thank you.
 
  • #6
In what proportion should the temperature rise in order to increase both volume and pressure c times?

Reject that the amount of gas is 1 mol. Assume instead, that the gas is at temperature Ti initially, and Pi=1 atm, Vi=1 L. Changing the temperature, the new pressure is P=cPi, the new volume is V=cVi. Express T in terms of c and Ti.

As for the set-up, that spring is a good idea, but the spring alone would not provide a pressure directly proportional to the volume, as the spring force is proportional to the change of volume, ΔV. But you can assume that the piston is also subjected to the atmospheric pressure, 1 atm. Elaborate.

ehild
 
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  • #7
I've worked it out prior to this and I believe from (PV/T)i=(PV/T)f

Tf= c2Ti, when Pf= cPi and Vf= cVi

Does that imply that dT must increase in intervals of c2 in relation to dTi?

-----
If I am understanding the statement correctly, the setup with a spring (of initial length when PV=0) would be valid, if and only if, the initial conditions is such that
-kx/A= P= n atm and V= n L, where n= 0, 1, 2, 3, ... n

Am I going off course here?

-----
Thank you.
 
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  • #8
CoinToss said:
I've worked it out prior to this and I believe from (PV/T)i=(PV/T)f

Tf= c2Ti, when Pf= cPi and Vf= cVi

Does that imply that dT must increase in intervals of c2 in relation to dTi?

dT is the increment of the temperature. Nothing is said how the variables has to change in time.

-----
CoinToss said:
If I am understanding the statement correctly, the setup with a spring (of initial length when PV=0) would be valid, if and only if, the initial conditions is such that
-kx/A= P= n atm and V= n L, where n= 0, 1, 2, 3, ... n

Am I going off course here?
.

No, you even have to go further. Design the experimental setup in detail.

Theoretically it is correct to assume that P=-kx/A but take care of the units.

But it is not realistic to assume that PV=0. That would mean either zero pressure or zero volume of the gas when the spring is relaxed. So the spring has to be compressed initially, when P=1 atm and the gas occupies 1 L volume.
You also have to take the environment into account. P=-kx/A is valid for zero external pressure. So you need to put cylinder into a vacuum chamber. Is not it easier to place it on the table in the lab or classroom? There is air outside, pressure very near to 1 atm, pressing the piston. In equilibrium, the pressure of the gas is the same as the pressure outside. So you can choose the initial condition of the spring that it is relaxed when the pressure of the gas is 1 atm and the volume 1 L. Taking the external pressure Pe into account, P=Pe-(kx)/A, V=Vi-Ax (x is negative as the spring is compressed). Initially, Pi=Pe =1 atm =105Pa and Vi=1L=10-3m3.
Elaborate how the spring constant has to be related to the size of the cylinder and the area of the piston.
Do the same in your setup, when the pressure of the gas is balanced entirely by the spring.

Assume that you have got a spring with k=900 N/m. What should be the diameter of the piston and he length of the cylinder?

ehild
 
  • #9
Hello ehild,

I feel awful for forgetting to thank you for all the contribution you've made towards this question, thank you. You've made me realize the alternative solution of not factoring out environmental effects on a system and I think that would be very valuable to me in the future.
As for the homework itself, it was due so I had to pass it up; turns out the trivial answer was, as your previous posts hinted, to add a specific amount of heat to the system. However, I was not penalized for my solution.

Again, thank you; I hope to have the privilege of your guidance again.
 
  • #10
The set-up you suggested was a very valuable one, and it ensured that the volume and pressure increase proportionally when you heat up the gas, while the process is quasi-static.
The heat adsorbed by the gas depends on the way how you rich the final state from the initial one. If you want to increase both the volume and pressure three times the original values, no matter on what way, the heat would be different from that adsorbed during your "CoinToss" process. Try to calculate the heat when the gas (1 mol oxygen, for example) expands at constant pressure of 1 atm from 1 L to 3 L and then, while the volume kept constant, the pressure increases to 3 atm by heating the gas. And calculate also the heat when p=(1atm/1 L )*V during the whole process.

ehild
 

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the relationships between heat, work, energy, and temperature, and how they affect physical systems.

2. What is the relationship between P and V in thermodynamics?

In thermodynamics, P (pressure) and V (volume) are related by the ideal gas law, which states that as one of these values increases, the other must decrease in order to maintain a constant temperature.

3. How does a process cause both P and V to increase in thermodynamics?

In thermodynamics, a process that causes both P and V to increase is known as an isentropic process. This occurs when a system undergoes a change in pressure and volume while maintaining a constant entropy.

4. What is the significance of P and V increasing simultaneously in thermodynamics?

When P and V increase simultaneously in a thermodynamic process, it means that the system is expanding and doing work on its surroundings, while also having an increase in internal energy. This is an important concept in understanding energy transfer and work done in physical systems.

5. Can you give an example of a process that causes both P and V to increase in thermodynamics?

An example of a process that causes both P and V to increase in thermodynamics is the expansion of a gas in a piston-cylinder system. As the gas expands, its volume increases and it pushes against the piston, causing an increase in pressure. This process also involves an increase in internal energy due to the work done by the expanding gas.

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