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You would need to find the National ES and ask him yourself. Each ES is different.

TheChiScientist wrote:You would need to find the National ES and ask him yourself. Each ES is different.

Have they released any info of who the National ES is going to be?

Tbh not that I am aware of. I would PM Unome as he is more aware of this kind of stuff.

Vrund wrote:

TheChiScientist wrote:You would need to find the National ES and ask him yourself. Each ES is different.

Have they released any info of who the National ES is going to be?

Maybe also PM chalker. He is a nat'l supervisor and he's on the physics commitee.

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.

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.

Hint: Adiabatic processes are polytropic.

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

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.

Hint: Adiabatic processes are polytropic.

Actually though, how do you solve that?

PV=nRT and U=3/2nRT

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

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?

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