Question involving resistance and heat transfer

[pc2-brazil]

Homework Statement

A resistance, connected to a battery, is put inside a thermally insulated cylinder which has an adjusted frictionless piston and contains an ideal gas. Through the resistance, a current i = 240 mA flows. At which velocity v should the piston move upwards, so that the temperature of the gas remains constant? The value of the resistance is R = 550 ohms and the mass of the piston is 11.8 kg.
[Note: the attached picture is scanned from "Physics" by Halliday, Resnick and Krane, 4th edition, Brazillian Portuguese version.]

Homework Equations

Joule effect:
P=Ri^2

The Attempt at a Solution

P=Ri^2 = (550)(240\times 10^{-3}) = 132 W
So, if this energy is entirely transferred to the gas, then the gas receives 132 J at each second.
If the volume remained constant, the temperature of the gas would increase. So, the piston should move upwards, thus allowing the gas to expand, increasing its volume and keeping the temperature constant.
I'm not sure if I understand this situation properly. I think that the heat Q transferred to the gas at each second equals 132 J. This heat will equal the work done by the gas against the piston to expand, maintaining the temperature constant.
I'm not sure how to proceed from here. I also don't know how the weight of the piston will interfere in the process.

Thank you in advance.

 

[Andrew Mason]

You have to square the current.

The voltage is 132 Volts. The current is .24A so the power is VI = 31.7 Joules/sec or watts.

So in one second:
Q = 31.7 J
ΔU = 0 (assuming an ideal gas)
W = mgΔh

Since the speed is constant there is no change in kinetic energy.

Does that help?

AM

 

[pc2-brazil]

You have to square the current.

The voltage is 132 Volts. The current is .24A so the power is VI = 31.7 Joules/sec or watts.

So in one second:
Q = 31.7 J
ΔU = 0 (assuming an ideal gas)
W = mgΔh

Since the speed is constant there is no change in kinetic energy.

Does that help?

AM

Yes, that helps.
So, I had the values of Q (apart from not having squared the current) and ΔU; I was only missing the value of W.
It will give:
Q = ΔU + W
31.7 = 0 + (11.8)(9.81)Δh
Δh = 27.4 cm
Because Δh is the displacement at each second, it corresponds to the velocity v of the piston, in cm/s. [This is the correct answer indicated in the book.]

So, the forces acting on the piston are its weight (mg) downwards, the force due to atmospheric pressure (p_atmA, where A is the area of the piston) downwards and the force due to the pressure of the gas (pA), upwards.
But what is keeping the piston's velocity constant? Is it a force of magnitude |mg + p_atmA - pA|?

 

[Andrew Mason]

But what is keeping the piston's velocity constant? Is it a force of magnitude |mg + p_atmA - pA|?

If it is moving at constant speed is there a net force acting on it? (hint: Newton's first law).

AM

 

[I like Serena]

I don't get it.
Isn't the work done = (mg + p_atmA)Δh?

That would either mean we have insufficient information, or we need to assume that an external force is be applied equal to p_atmA...

 

[Andrew Mason]

I don't get it.
Isn't the work done = (mg + p_atmA)Δh?

That would either mean we have insufficient information, or we need to assume that an external force is be applied equal to p_atmA...

If the top of the piston was exposed to atmospheric pressure, you would be correct, of course. But there is no indication of that in the question or drawing. There is just an ideal gas below. Also, area is not given so we are to assume it is immaterial - ie. no gas pressure downward on the piston.

AM

 

[pc2-brazil]

If it is moving at constant speed is there a net force acting on it? (hint: Newton's first law).

AM

There is certainly no net force. Since there is no net force, this means that there must be a force of magnitude |mg + p_atmA - pA|, applied by an external agent, which opposes the net force on the piston due to the pressure of the gas, atmospheric pressure and gravity. Is this correct?