# Change in internal energy of a cylinder

• guyvsdcsniper
In summary, the system does some work on the environment and some work internally which is not counted in the equation. So by knowing Q=0 and that work is negative, I can determine that whatever answer I get for work, I need to tack on a negative sign.
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?

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

Doc Al said:
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?

quittingthecult said:
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.

Doc Al said:
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.

quittingthecult said:
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.)
quittingthecult said:
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.

Doc Al said:
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?

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
Doc Al said:
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

## 1. What is the definition of change in internal energy of a cylinder?

The change in internal energy of a cylinder refers to the difference between the initial and final internal energies of the system. It takes into account any changes in the temperature, pressure, and volume of the gas inside the cylinder.

## 2. How is the change in internal energy of a cylinder calculated?

The change in internal energy can be calculated using the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system. This can be represented by the equation: ΔU = Q - W.

## 3. What factors affect the change in internal energy of a cylinder?

The change in internal energy of a cylinder is affected by the amount of heat added or removed from the system, the work done by the system, and the type of gas inside the cylinder. Changes in temperature, pressure, and volume also play a role in determining the change in internal energy.

## 4. How does the change in internal energy of a cylinder relate to the ideal gas law?

The ideal gas law, which states that the pressure, volume, and temperature of an ideal gas are all directly proportional, can be used to calculate the change in internal energy of a cylinder. By manipulating the ideal gas law equation, it is possible to determine the change in internal energy of a gas inside a cylinder.

## 5. What are some practical applications of understanding the change in internal energy of a cylinder?

Understanding the change in internal energy of a cylinder is important in many industrial and scientific applications. It is used in the design and operation of engines, refrigeration systems, and other thermodynamic processes. It also plays a role in the study of chemical reactions and the behavior of gases under different conditions.

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