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jinhusong wrote:

Alex-RCHS wrote:

jinhusong wrote:PV=nRT and U=3/2nRT

The temperature raising because of both V and work (forceXdistance).

I'm afraid I still don't understand. How do you find U? I could see doing force times distance but I don't see any way to find the distance over which the force acts (the amount the piston descends).

And as for the ideal gas law, you don't know what V would be, right?

I did not check the calculation and units. Here is an idea:

Piston area: A=3.14 * 0.125 * 0.125 = 0.0491 (m*m)

Initial:
P0=4.5X9.8/A=898.2 N/(m*m)

V0=n*R*T0/P0=8 * 8.314459 * 278 / 898.2 = 20.59 (m*m*m)

h0 = V0 / A = 419.3m

U0=3/2 * R * T0 * 8 = 27737

======
After apply force of 50N

P2=(4.5X9.8 + 50)/A= 1916.5


U2 = 3 /2 * R * 8 * T2

U2 = U0 + (4.5X9.8 + 50) * LLLL
LLLL is the distance the piston moved.

First equation
3/2 * 8.31446 * 8 * T2 = 27737 + 94.1 * LLLL


======
P2 X (h0-LLLL)*A=8* R * T2


Second equation
1916.5*(419.3-LLLL)*0.0491= 8 * 8.31446 * T2



(1)
LLLL = 1.060 * T - 294.76


(2)
LLLL = 419.3 - 0.7069 * T


T= 404K

You are a beast, thanks.

Seeing division C get college level questions makes me wonder; will division B get college level questions, or will we just get AP Phys. II level questions? (looking at the BDJH B division test that we perfect scored, the BDJH C test makes it look like a midget)

JoeyC wrote:

jinhusong wrote:

Alex-RCHS wrote: I'm afraid I still don't understand. How do you find U? I could see doing force times distance but I don't see any way to find the distance over which the force acts (the amount the piston descends).

And as for the ideal gas law, you don't know what V would be, right?

I did not check the calculation and units. Here is an idea:

Piston area: A=3.14 * 0.125 * 0.125 = 0.0491 (m*m)

Initial:
P0=4.5X9.8/A=898.2 N/(m*m)

V0=n*R*T0/P0=8 * 8.314459 * 278 / 898.2 = 20.59 (m*m*m)

h0 = V0 / A = 419.3m

U0=3/2 * R * T0 * 8 = 27737

======
After apply force of 50N

P2=(4.5X9.8 + 50)/A= 1916.5


U2 = 3 /2 * R * 8 * T2

U2 = U0 + (4.5X9.8 + 50) * LLLL
LLLL is the distance the piston moved.

First equation
3/2 * 8.31446 * 8 * T2 = 27737 + 94.1 * LLLL


======
P2 X (h0-LLLL)*A=8* R * T2


Second equation
1916.5*(419.3-LLLL)*0.0491= 8 * 8.31446 * T2



(1)
LLLL = 1.060 * T - 294.76


(2)
LLLL = 419.3 - 0.7069 * T


T= 404K

You are a beast, thanks.

Wow, thank you.

jinhusong wrote:

Alex-RCHS wrote:

jinhusong wrote:PV=nRT and U=3/2nRT

The temperature raising because of both V and work (forceXdistance).

I'm afraid I still don't understand. How do you find U? I could see doing force times distance but I don't see any way to find the distance over which the force acts (the amount the piston descends).

And as for the ideal gas law, you don't know what V would be, right?

I did not check the calculation and units. Here is an idea:

Piston area: A=3.14 * 0.125 * 0.125 = 0.0491 (m*m)

Initial:
P0=4.5X9.8/A=898.2 N/(m*m)

V0=n*R*T0/P0=8 * 8.314459 * 278 / 898.2 = 20.59 (m*m*m)

h0 = V0 / A = 419.3m

U0=3/2 * R * T0 * 8 = 27737

======
After apply force of 50N

P2=(4.5X9.8 + 50)/A= 1916.5


U2 = 3 /2 * R * 8 * T2

U2 = U0 + (4.5X9.8 + 50) * LLLL
LLLL is the distance the piston moved.

First equation
3/2 * 8.31446 * 8 * T2 = 27737 + 94.1 * LLLL


======
P2 X (h0-LLLL)*A=8* R * T2


Second equation
1916.5*(419.3-LLLL)*0.0491= 8 * 8.31446 * T2



(1)
LLLL = 1.060 * T - 294.76


(2)
LLLL = 419.3 - 0.7069 * T


T= 404K

But why would the initial pressure be just the weight of the gas divided by the area if the particles are in motion?

UTF-8 U+6211 U+662F wrote:

jinhusong wrote:

Alex-RCHS wrote: I'm afraid I still don't understand. How do you find U? I could see doing force times distance but I don't see any way to find the distance over which the force acts (the amount the piston descends).

And as for the ideal gas law, you don't know what V would be, right?

I did not check the calculation and units. Here is an idea:

Piston area: A=3.14 * 0.125 * 0.125 = 0.0491 (m*m)

Initial:
P0=4.5X9.8/A=898.2 N/(m*m)

V0=n*R*T0/P0=8 * 8.314459 * 278 / 898.2 = 20.59 (m*m*m)

h0 = V0 / A = 419.3m

U0=3/2 * R * T0 * 8 = 27737

======
After apply force of 50N

P2=(4.5X9.8 + 50)/A= 1916.5


U2 = 3 /2 * R * 8 * T2

U2 = U0 + (4.5X9.8 + 50) * LLLL
LLLL is the distance the piston moved.

First equation
3/2 * 8.31446 * 8 * T2 = 27737 + 94.1 * LLLL


======
P2 X (h0-LLLL)*A=8* R * T2


Second equation
1916.5*(419.3-LLLL)*0.0491= 8 * 8.31446 * T2



(1)
LLLL = 1.060 * T - 294.76


(2)
LLLL = 419.3 - 0.7069 * T


T= 404K

But why would the initial pressure be just the weight of the gas divided by the area if the particles are in motion?

Ah, this I actually know (kinematics/statics is easy for me but I'm not too good at thermo).

It's because the force with which the gas pushes up on the piston is equal to the force of gravity pushing down on the piston, because the piston is not moving. And it wouldn't be the weight of the gas, it would be the weight of the piston itself (or the top part of it, at least).

Oh, whoops, misread the question haha

So..... Problem resolved?

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)

Hum. Makes sense. Thanks.

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.