Change in internal energy of a cylinder

[guyvsdcsniper]

Homework Statement

4.6 Friction / Dissipation. A volume of gas is contained in a wellinsulated
cylinder with a well- insulated piston head as depicted in Figure
4.4. The massless piston head may move but only by overcoming 0.2 newtons
of kinetic friction. A 50- g mass is placed on top of the piston head.
The piston head moves outward a distance of 25 cm. Ignore the pressure
exerted by the atmosphere.

(a) What is the amount of work performed by the gas during this
expansion?

(b) If we consider the gas, the cylinder, and the piston head to be part
of the system, what is the change in internal energy, ΔE, of the
system?

Relevant Equations

ΔE=Q+W

I believe I got the first part of this questions solved.

For part b, we are asked to find the change in internal energy.

We know ΔE=Q+W. The cylinder,gas and piston head are the system. The cylinder and piston head are well insulated, so there will be no head transfer, therefore Q=0.

So now we have ΔE=W. From part a, we found the work done to be .1725J.
By the equation above the change in energy is .1725J.

I don't know if that answer makes sense to me. The system is doing the work and by convention, work done by system is W<0.

But here I am getting a positive answer for overall change in energy.

Am I getting something wrong here?

 

[Doc Al]

Careful with signs. In your first law equation, the W is work done on the system.

 

[guyvsdcsniper]

Careful with signs. In your first law equation, the W is work done on the system.

Ok so in part b the overall system doing work. A system doing work means W<0. So I should use ΔE=Q-W because I know the system is doing work?

 

[Doc Al]

Ok so in part b the overall system doing work. A system doing work means W<0. So I should use ΔE=Q-W because I know the system is doing work?

Careful. Some of that work is on the "environment" and thus will be negative in that equation. But a portion of the total work done is purely internal to the system and thus doesn't change the internal energy.

 

[guyvsdcsniper]

Careful. Some of that work is on the "environment" and thus will be negative in that equation. But a portion of the total work done is purely internal to the system and thus doesn't change the internal energy.

So just by knowing that some of the work is on the enviornment, that should just tell me that I need to change ΔE=Q+W to ΔE=Q-W?

I looked up the solution to this problem and the solver used ΔQ= ΔE+W and then got ΔE=-W. Could you help me understand how this equation can be arranged like this? I don't see how I can get to that equation form ΔE=Q+W.

 

[Doc Al]

So just by knowing that some of the work is on the enviornment, that should just tell me that I need to change ΔE=Q+W to ΔE=Q-W?

When you use ΔE=Q+W, work done on the system will be positive but work done by the system will be negative. Here it's the system doing work, so W will be negative. (But... some of the work done by the expanding gas is purely internal to the system, so you don't count it in this equation.)

I don't see how I can get to that equation form ΔE=Q+W.

They are just playing around with signs (like I suggest above), no deep analysis here.

 

[guyvsdcsniper]

When you use ΔE=Q+W, work done on the system will be positive but work done by the system will be negative. Here it's the system doing work, so W will be negative. (But... some of the work done by the expanding gas is purely internal to the system, so you don't count it in this equation.)

They are just playing around with signs (like I suggest above), no deep analysis here.

Got it. I just want to make sure I am understand this correctly.

The system is doing work on the environment, but there is also some work done internally but since it is internal I do not need to include this into my equation

So with ΔE=Q+W, I know Q=0 so ΔE=W.

Now by analyzing the problem, I know work was done by the system and so that tells me work is negative (omitting the internal work). There for I should know that what ever answer I get for work, I need to tack on a negative sign to it?

 

[Doc Al]

Yes, the work done is counted as negative, since the system is doing the work on the environment.

If you look at it from the environment's point of view it might be more obvious: The environment is exerting a force inward (the weight of the mass) but the displacement is outward. Thus the work done by that force must be negative.

 

[guyvsdcsniper]

Yes, the work done is counted as negative, since the system is doing the work on the environment.

If you look at it from the environment's point of view it might be more obvious: The environment is exerting a force inward (the weight of the mass) but the displacement is outward. Thus the work done by that force must be negative.

Thank you for the help again. I greatly appreciate it