First Law of Thermodynamics - Piston

1. Nov 27, 2015

rpthomps

A cylinder of cross-sectional Area A is closed by a massless piston. The cylinder contains n mol of ideal gas with specific heat ratio and is initially in equilibrium with the surrounding air at temperature T and pressure P. The piston is initially at height h1 above the bottom of the cylinder. Sand is gradually sprinkled onto the piston until it has moved downward to a final height h2. Find the total mass of the sand if the process is isothermal

2. Relevant equations

W=PV

W=mgh

3. The attempt at a solution

$\Delta U=\Delta Q-W\\\\$

But the system is isothermal therefore

$\Delta U= 0\\\\\Delta Q =W$

The Work is being done by gravity

$mg\Delta h=PV\\\\mg(h_2-h_1)=PA(h_2)\\\\m=\frac{PA(h_2)}{(h_2-h_1)g}$

$M=\frac{PA}{g}(\frac{h_1}{h_2}-1)$

2. Nov 27, 2015

Staff: Mentor

This is not a 1st law problem. This is an ideal gas law problem.

3. Nov 27, 2015

Sturk200

4. Nov 27, 2015

rpthomps

Yes, I agree. I thought I had kept that concept in mind in my solution above. Is this not the case?

Thanks again for looking.

5. Nov 27, 2015

Staff: Mentor

No. Your analysis is incorrect. It doesn't even determine the work correctly. Furthermore, the work is irrelevant to the solution of this problem.

What is the initial pressure and the initial volume? What is the final volume? From the ideal gas law, what is the final pressure? Considering the weight of the sand and the initial pressure, what is the final pressure? What do you get if you set your two relationships for the final pressure equal to one another?

6. Nov 27, 2015

Sturk200

Not quite. Isothermal means that temperature is held constant, whereas adiabatic means that no heat (thermal energy) is exchanged. (Remember that U stands for thermal energy.) If you look at the etymology, adiabatic means literally "not go through," so it refers to some physically intrinsic thing (energy) that is not being allowed to pass a barrier. Isothermal just means "equal temperature" -- no mention of a barrier. Thus energy is allowed to be exchanged in an isothermal process, and indeed the exchange of energy during such a process is often precisely the thing that allows the system to be kept at the same temperature throughout.

7. Nov 28, 2015

rpthomps

Okay..I will look at the problem again but before I do, doesn't the statement $\Delta U=0$ mean that the gas doesn't increase its temperature and that by making $\Delta Q$ not 0 that I am allowing for the possibility of an adiabatic transfer of energy?

Again, thanks for spending your time on this.

8. Nov 28, 2015

Sturk200

Nope, setting U=0 means that you are not allowing any heat energy to enter or leave the system. Under this assumption, if we add mechanical work to the system, for instance, then Q would have to go up and thus the temperature would have to increase. If, on the other hand, we had allowed U to be nonzero, then the input of mechanical work may have been counterbalanced by an output of heat energy, thus allowing the overall Q (and thus temperature) to remain constant.

9. Nov 28, 2015

Staff: Mentor

This is totally incorrect. For an ideal gas, if ΔU = 0, the temperature is constant. Q = 0 (not ΔU=0) means that you are not allowing any heat energy to enter or leave the system. So, I agree with post #7. Not that this has anything to do with the solution of this problem, which is strictly an ideal gas law application.

Chet

10. Nov 28, 2015

Sturk200

Woah, yes that's completely right. Sorry. The book I use has a plus sign convention for W and I think that threw me off.

11. Nov 28, 2015

Staff: Mentor

Chet

12. Nov 28, 2015

rpthomps

Hey Chet: Still working on it. :) I usually have to think quite a bit about this stuff. What I lack in natural talent though, I gain in perseverance.

13. Nov 28, 2015

Mister T

For an ideal gas, $\Delta U=0$ means the temperature is constant.

$\Delta Q=0$ means the process is adiabatic. Making $\Delta Q$ nonzero means there is a transfer of heat energy.

14. Nov 29, 2015

rpthomps

USEFUL EQUATIONS:
$P_iV_i=nRT\\\\P_fV_f=nRT\\\\V_i=h_1A\\\\V_f=h_2A\\\\$

ATTEMPT AT A SOLUTION:

through substation and elimination

$P_f=P_i\frac{h_1}{h_2}$

At h2 mass of sand is in static equilibrium with the pressure of the gas, using Newton’s equations:

$P_fA=mg\\\\P_f=\frac{mg}{A}\\$

More substitutions:

$m=\frac{P_ih_1A}{gh_2}$

Which is different then the book still. Not sure, how they got the difference in values, that’s why I was thinking work originally.

15. Nov 29, 2015

Staff: Mentor

Your answer for Pf in terms of mg is incorrect. You left something out. What is it that you think you might have omitted from the equation?

Chet

16. Nov 29, 2015

Mister T

Right. This follows from Boyle's Law.

I made this same mistake when I first did it in my head. I had to write it out before I realized that $\frac{mg}{A}$ is the pressure increase (caused by the sand).

17. Nov 29, 2015

Staff: Mentor

Don't you think it would have been better for rpthomps if he had been given the opportunity to deduce this himself?

18. Nov 29, 2015

rpthomps

Is there some kind of normal force that is present before the sand is on? This would be equivalent to $P_i\times A$ The sand adds mg to this downward force which is balanced by $P_f\times A$. Solving for m gives me the expression at the back of the text. I have a real difficulty seeing this. I just assumed the weight of the sand was equivalent to the pressure of the gas. Thanks again for all your help!

19. Nov 29, 2015

Staff: Mentor

The words "initially in equilibrium with the surrounding air at temperature T and pressure P" in the problem statement means that the piston is initially in force equilibrium. Do a force balance on the piston before the sand is added, starting with a free body diagram. What are the forces acting on the massless piston initially? What is the force balance equation on the massless piston initially?

Chet

20. Nov 29, 2015

rpthomps

Oh, it is the pressure of the air above it! :) Thank you so much sir!