Isothermal Compression of a Ideal Gas and Distance

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Homework Help Overview

The problem involves the isothermal compression of an ideal gas within a cylindrical tank, where a movable disk compresses the gas as weight is added. The initial height of the disk is given, along with the masses of the disk and an additional weight. The participants are tasked with determining the final position of the disk under these conditions, while maintaining a constant temperature for the gas.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between pressure and force, questioning how to relate the initial and final pressures without specific volume or area values. Some suggest that the area may cancel out in calculations.
  • There are algebraic inquiries into the forces and pressures acting on the disk, with participants attempting to express these in terms of the known variables.
  • Some participants express confusion about the role of atmospheric pressure in the calculations and whether it should be included in their analysis.

Discussion Status

The discussion is ongoing, with participants providing algebraic expressions for the forces and pressures involved. There is a recognition of the complexity introduced by atmospheric pressure, and some guidance has been offered regarding the assumptions made in the initial analysis. However, no consensus has been reached on how to proceed without additional information.

Contextual Notes

Participants note the absence of specific values for initial and final pressures, as well as the area of the disk, which complicates their ability to find a definitive solution. The assumption of atmospheric pressure versus a vacuum outside the cylinder is also under consideration.

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Homework Statement


The figure at the bottom shows a cylindrical tank of diameter D with a moveable 3.00-kg circular disk sitting on top. The disk seals in the gas inside but is able to move without friction. The gas inside is at temperature T. The height of the disk is initially at h = 4.00 m above the bottom of the tank. When a lead brick of mass 9.00 kg is gently placed on the disk, it moves downward, compressing the gas and decreasing the volume. Assuming the temperature of the gas is held constant, at what distance from the bottom of the will the disk come to rest?

media%2F38a%2F38ae8a2b-f40e-4de5-a308-1866746910bb%2Fphp3V4qiK.png


mass_{disk}=3.00 kg
mass_{Pb}=9.00 kg
m_{T}=12.00kg
h_i=4.00m
Isothermal \therefore T_i=T_f
V_i > V_f ; P_i<P_f

Homework Equations


\P = \rho*\g*\h
P_1*V_1*T_2=P_2*V_2*T_1
W=nRT*ln(\frac{V_f}{V_i})=m*g*d

The Attempt at a Solution



\[/B]delta P = -\rho*g*\int_{y}^{4} = -0.5*\rho*g*(y^2-16)
F_{disk+Pb}=(m_{d}+m{Pb})/g=127.53N
P_1*V_1=P_2*V_2
P_1*A_1*H_1=P_2*A_2*H_2 (A_1=A_2)
H_2=4*\frac{P_1}{P_2}

I'm not really sure where to proceed here since the initial pressure nor the final pressure nor any of the volumes are given so I'm generally extremely confused on how to proceed to find the final distance the piston would rest at. I know since that by Boyles law as Volume decreases, Pressure increases at constant Temperature. I'm just not sure how we factor any of that in without being given volume or area.
 

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I haven't spent the time to work out the solution but I note that Pressure = Force/Area so my guess would be that the area cancels at some point.
 
Algebraically, in terms of mdisk and g, what is the force of the gas on the disk initially? Algebraically, in terms of mdisk, g, and A (where A is the area of the disk), what is the pressure of the gas P1 on the disk initially?

Algebraically, in terms of mdisk, mPb, and g, what is the force of the gas on the disk finally? Algebraically, in terms of mdisk, mPb, g, and A (where A is the area of the disk), what is the pressure of the gas P2 on the disk finally?

Algebraically, what is the ratio of P1 to P2?
 
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I've got this but I'm not sure how to proceed from here without being given any of the pressures.

$$\delta P = -\rho*g*\int_{y}^{4} = -0.5*\rho*g*(y^2-16)$$
$$ F_{disk+Pb}=(m_{d}+m{Pb})/g=127.53N $$
$$ P_1*V_1=P_2*V_2 $$
$$ P_1*A_1*H_1=P_2*A_2*H_2 (A_1=A_2) $$
$$H_2=4*\frac{P_1}{P_2}$$
 
elements said:
I've got this but I'm not sure how to proceed from here without being given any of the pressures.

$$\delta P = -\rho*g*\int_{y}^{4} = -0.5*\rho*g*(y^2-16)$$
$$ F_{disk+Pb}=(m_{d}+m{Pb})/g=127.53N $$
$$ P_1*V_1=P_2*V_2 $$
$$ P_1*A_1*H_1=P_2*A_2*H_2 (A_1=A_2) $$
$$H_2=4*\frac{P_1}{P_2}$$
You haven't answered my questions.
 
Chestermiller said:
Algebraically, in terms of mdisk and g, what is the force of the gas on the disk initially? Algebraically, in terms of mdisk, g, and A (where A is the area of the disk), what is the pressure of the gas P1 on the disk initially?Algebraically, in terms of mdisk, mPb, and g, what is the force of the gas on the disk finally? Algebraically, in terms of mdisk, mPb, g, and A (where A is the area of the disk), what is the pressure of the gas P2 on the disk finally?

Algebraically, what is the ratio of P1 to P2?

So the initial Force is described algebraically as
Fg=mdisk*g
Initial Pressure
P1=F/A=mdisk*g/A
P2=mdisk+Pb*g/A
P1/P2=V2/V1=A*H2/A*H1

mdisk*g*A-1/mdisk+Pb*g*A-1= 4(mDisk/mPb+Disk)=H2

Is that how height gets related to pressure? What about atmospheric pressure or does that not factor in?
 
elements said:
So the initial Force is described algebraically as
Fg=mdisk*g
Initial Pressure
P1=F/A=mdisk*g/A
P2=mdisk+Pb*g/A
P1/P2=V2/V1=A*H2/A*H1

mdisk*g*A-1/mdisk+Pb*g*A-1= 4(mDisk/mPb+Disk)=H2

Is that how height gets related to pressure? What about atmospheric pressure or does that not factor in?
VERY NICELY DONE!

The analysis you have done so far assumes that the outside pressure is zero (i.e., vacuum). Let's go back to the equilibrium force balances on the piston, and repeat them assuming that the outside pressure is atmospheric (patm). Again, please do this algebraically. What do you get for the inside gas pressures then?
 
Chestermiller said:
VERY NICELY DONE!

The analysis you have done so far assumes that the outside pressure is zero (i.e., vacuum). Let's go back to the equilibrium force balances on the piston, and repeat them assuming that the outside pressure is atmospheric (patm). Again, please do this algebraically. What do you get for the inside gas pressures then?

P1=F/A=mdisk*g/A + Patm
P2=mdisk+Pb*g/A + Patm
P1/P2=mdisk*g/A + Patmmdisk+Pb*g/A + PatmThe areas no longer cancel out here so I am guessing the atmospheric pressures cancel somehow?
 
elements said:
P1=F/A=mdisk*g/A + Patm
P2=mdisk+Pb*g/A + Patm
P1/P2=mdisk*g/A + Patmmdisk+Pb*g/A + PatmThe areas no longer cancel out here so I am guessing the atmospheric pressures cancel somehow?
No. I guess they expected you to assume vacuum outside the cylinder. But you did great on the case with atmospheric pressure, especially realizing that you need to know A to get a solution.
 

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