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ScienceTurtle314 wrote:

jinhusong wrote:

TheChiScientist wrote:So..... Problem resolved?

I feel this is a test question, so they will not expect you take that long time to do the number calculation. If you keep the symbols, and do not replace them with numbers, you will get:

from U1=U0+(mg+F)(V0-V1)/A and U0=3/2nRT0, U1=3/2nRT1 you got: T1-T0=(mg+F) * 2/3 * (V0-V1)/(nRA), where A is the piston area.

from P0V0=nRT0 and P1V1=nRT1, you got V0-V1=nR(T0/P0 - T1/P1)

replace the V0-V1 in the first equation and remember P0A=mg, and P1A=mg+F, you get
(T1-T0)=2/5 * F/(mg) * T0. Only decided by initial temperature, weight of piston and force applied.

So the answer to the question (T1-T0)=2/5 * 50 / (4.5*9.8) * 278 = 126F
(T1 = 278 + 126 = 404F)

This was a good problem, and this is a good solution attempt. There's a problem though with how you find that the internal energy of the system when the work done by the piston (energy lost) F∆d is equal to the energy gained by the gas 3/2nR∆T. This implies that all of the energy from the pressure of the piston goes into the internal energy of the gas. However, in reality when the force of the piston on the gas is in equilibrium with the force of the gas on the piston, the piston will be in motion and so it will have kinetic energy, so F∆d is actually less than 3/2nR∆T because energy goes into the kinetic energy of the piston. To get this, think about about what happens when the piston first falls into gas and compresses it. When the force of the piston on the gas (F) reaches the force of the gas on the piston (Pressure*A), the piston will stop accelerating downwards (technically the forces on the piston and gas are equal by Newton's 3rd law, but I mean the force from the pressure of the piston and the force from the pressure of the gas). However, it will still have velocity from its acceleration up to that point, and therefore continue to move downwards, so that the gas's force upwards pressure is greater than the downwards pressure from the piston. The resulting upwards acceleration will cause the piston to come to a rest, and then begin to accelerate back upwards towards the force equilibrium. Over time, the piston should actually oscillate back and forth around the force equilibrium, such that it always is in motion at that point. We cannot say that F∆d is equal to 3/2nR∆T. When solving this problem properly, you would instead find work when forces are at equilibrium using dU = PdV = 3/2nRdT, and throwing in an additional equation for adiabatic compression (dU = [R/(1-gamma)]dT). These can be used to calculate dU, and circumvent the need to calculate how far the piston falls. Because all the math is complex (and beyond my current knowledge) and simplifies well, the formula you can use to solve this problem is that P1^(1−γ1)*T1^γ=P2^(1−γ2)*T2^γ2... if you google free expansion you can find out more about it. Using this, you get a change of 98K, which is the solution to the problem given by the key in the original question.

jinhusong wrote: I am wrong. Thanks for the right answer. The equation is P1^(1-gamma)T1^gamma)=P2^(1-gamma)T2^gamma. gamma=5/3 for this case.
I got into a wrong path.

Here are something still bother me.

My feeling is you apply the force gradually, extremely slow, from 0 to 50N and the piston no move once it reach 50N.

If you really apply a constant 50N force, it will end up like you said, piston oscillation and temperature up and down, non-stop.

ScienceTurtle314 wrote:

jinhusong wrote: I am wrong. Thanks for the right answer. The equation is P1^(1-gamma)T1^gamma)=P2^(1-gamma)T2^gamma. gamma=5/3 for this case.
I got into a wrong path.

Here are something still bother me.

My feeling is you apply the force gradually, extremely slow, from 0 to 50N and the piston no move once it reach 50N.

If you really apply a constant 50N force, it will end up like you said, piston oscillation and temperature up and down, non-stop.

You could do something like that where you apply a constantly increasing force that is just enough so that the piston has barely any velocity at each point, so that at 50N the velocity would be negligible (but you still couldn't calculate work as force*distance as that only works with constant force... and to use
W = ∫Fdx doesn't make much sense because a formula for force would just lead you back to calculating the changing pressure of the gas as you would if you just started with adiabatic compression equations). Another way to make it work is to apply a force to the piston that is somewhat less than 50N, so that it oscillates down to a minimum height of where 50N of gas pressure is pushing up on it. At the minimum height you would then increase force on the piston to 50N, so that the forces are in equilibrium and the velocity is 0.

JoeyC wrote:HI guys! I just got the BDJH regional test, and looking at the one for C division I just can't figure out how to solve this.
1.)You have a piston filled with 8 moles of a monatomic ideal gas at a temperature of 278 K. It is completely insulated (meaning that there is no heat transfer between the walls of the piston) and it is at rest in a vacuum. The piston has a diameter of 25 cm and has a mass of 4.5 kg. You then apply 50 N of force to the top of the piston.

a. (8 points) Determine the change in temperature of the gas:

Answer:98K Does anyone know what formula they used, and how they used it; Ideal gas law doesn't work with both volume and temperature changing after compression.

You don't actually have to solve for internal energy or anything crazy like that. You can use the following adiabatic relationship to solve for everything else:

[math]P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma[/math]

P1 can be found by taking the mass of the lid, finding its weight, and dividing it by the area: [b]898.4 Pa[/b]
T1 is given: [b]278 K[/b]
P2 can be found by taking the force applied (50 N), dividing it by the area of the lid, and adding it to P1: [b]1917.0 Pa[/b]

The gas is monatomic which means that the value of gamma is 5/3. Plugging these numbers into the equation listed above and solving for T2 gives: [b]376.4 K[/b]

To solve for the change in temperature, just find the difference between T2 and T1, which gives: 98.4 K or [b]98 K[/b].

Alex-RCHS wrote:Quick update on the ice/no ice debate: between my prediction, my heat retention, and my ice bonus I received 39.88 points for my device at the state tournament (and that's even after the low volume [80ml] and low time [22 min] made it so that the benefit of the ice bonus was even smaller). Without the ice bonus, the max is 40 if you have the very best HRF and prediction. Unfortunately, I bombed the test and got 9th . This makes me quite confident that the Ice Bonus is in fact useful. My predictions have been consistently within half a degree of accuracy with the ice bonus, and at states I was off by only 0.4 degrees.

I think the top teams at nationals will all use the ice bonus.

5uper5tring wrote:

Alex-RCHS wrote:Quick update on the ice/no ice debate: between my prediction, my heat retention, and my ice bonus I received 39.88 points for my device at the state tournament (and that's even after the low volume [80ml] and low time [22 min] made it so that the benefit of the ice bonus was even smaller). Without the ice bonus, the max is 40 if you have the very best HRF and prediction. Unfortunately, I bombed the test and got 9th . This makes me quite confident that the Ice Bonus is in fact useful. My predictions have been consistently within half a degree of accuracy with the ice bonus, and at states I was off by only 0.4 degrees.

I think the top teams at nationals will all use the ice bonus.

What was the initial bath temperature? How much ice water did you add? Do you know if the top scoring team also added ice water?

Alex-RCHS wrote:

5uper5tring wrote:

Alex-RCHS wrote:Quick update on the ice/no ice debate: between my prediction, my heat retention, and my ice bonus I received 39.88 points for my device at the state tournament (and that's even after the low volume [80ml] and low time [22 min] made it so that the benefit of the ice bonus was even smaller). Without the ice bonus, the max is 40 if you have the very best HRF and prediction. Unfortunately, I bombed the test and got 9th . This makes me quite confident that the Ice Bonus is in fact useful. My predictions have been consistently within half a degree of accuracy with the ice bonus, and at states I was off by only 0.4 degrees.

I think the top teams at nationals will all use the ice bonus.

What was the initial bath temperature? How much ice water did you add? Do you know if the top scoring team also added ice water?

I don’t remember the initial temperature exactly, but it was about 75-85. We added a full 50ml. The winning team did not add ice water, but they won because of the test, not their device. (I spoke to them after and did the math to confirm this)

5uper5tring wrote:

Alex-RCHS wrote:

5uper5tring wrote:
What was the initial bath temperature? How much ice water did you add? Do you know if the top scoring team also added ice water?

I don’t remember the initial temperature exactly, but it was about 75-85. We added a full 50ml. The winning team did not add ice water, but they won because of the test, not their device. (I spoke to them after and did the math to confirm this)

Was the 39.88 the highest experimental score (prediction + heat retention + ice bonus) achieved in that event? If you had not added the ice, assuming your prediction was also within 0.5 deg without the ice, what do you think you score would have been?

Alex-RCHS wrote:

5uper5tring wrote:

Alex-RCHS wrote: I don’t remember the initial temperature exactly, but it was about 75-85. We added a full 50ml. The winning team did not add ice water, but they won because of the test, not their device. (I spoke to them after and did the math to confirm this)

Was the 39.88 the highest experimental score (prediction + heat retention + ice bonus) achieved in that event? If you had not added the ice, assuming your prediction was also within 0.5 deg without the ice, what do you think you score would have been?

I don’t know what the highest experimental score was, but for the teams that did not use ice the highest possible score is 40.

Based on my trials we would have gotten about 0.5-1.5 points lower without the ice bonus at this time, temp, and volume (I can’t be sure without knowing he best HRF, but I know it was beneficial). That’s just based on the ice and heat scores. I think our prediction would be made much less accurate without the ice water.