How Is Heat Transferred in Different Paths of a P-V Diagram Cycle?

AI Thread Summary
Heat transfer in a gas cycle depicted in a p-V diagram involves understanding the relationships between internal energy, work, and heat. For path ab, the gas does +5.3 J of work on the surroundings during expansion, leading to a net energy gain of +8.2 J when considering the overall internal energy change of +2.9 J. Path bc involves negative work due to compression, while path ac has no work done since there's no volume change. It's crucial to accurately define the quantities of heat (Q) and work (W) to avoid confusion and ensure correct calculations. Understanding these principles is essential for solving thermodynamic problems effectively.
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Homework Statement




A sample of gas is taken through the cycle abca shown in the p-V diagram of Fig. 19-9. The net work done is +1.2 J. Along path ab, the change in the internal energy is +2.9 J and the magnitude of the work done is 5.3 J. Along path ca, the energy transferred to the gas as heat is +2.8 J.
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(a) How much energy is transferred to the gas as heat along path ab?
(b) How much energy is transferred to the gas as heat along path bc?

Homework Equations



DU = W + Q


The Attempt at a Solution



I figured it might be 2.9 - 5.3 = -2.4 for part A.
But, its not the right answer...
 
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Along path ab, work is positive because the gas is expanding. When you got expansion, work is positive. Therfore, it can't be -5,3 it's +5,3.

Along path bc, work is negative (compression). Along path ac, there is no work done, because there is no change in the volume.
 
Azndoode1 said:

Homework Equations



DU = W + Q
You need to be clear on what the quantities Q and W stand for. Q is the amount of heat flowing into the gas. If heat flows out of the gas, Q would be negative. Similarly, W is the amount of work done on the gas by the surroundings. If the gas does work on the surroundings, W would be negative.

So for part (a), the gas expanded. So does it do 5.3 J of work on the surroundings or did the surroundings do 5.3 J of work on the gas?
 
The correct answer was actually +8.2...
 
method_man said:
Along path ab, work is positive because the gas is expanding. When you got expansion, work is positive. Therfore, it can't be -5,3 it's +5,3.

Along path bc, work is negative (compression). Along path ac, there is no work done, because there is no change in the volume.

vela said:
You need to be clear on what the quantities Q and W stand for. Q is the amount of heat flowing into the gas. If heat flows out of the gas, Q would be negative. Similarly, W is the amount of work done on the gas by the surroundings. If the gas does work on the surroundings, W would be negative.

You are being told two different things here. But it doesn't necessarily matter how you define your system as long as you can stay consistent.

As far as the answer 8.2 goes, that seems correct to me.
You have an isobaric expansion, and anytime something expands it does work on the surroundings. So for simplicity sake let's say that the gas had to sacrifice how compressed it was to do work on the surroundings. So your gas has just lost 5.3J.
But then it goes to tell you that the gas actually gained 2.9J overall. The only place it can gain energy is from heat. So how many Joules of heat did it gain to bring the overall internal energy increase to 2.9J?

Let me know if that makes sense.
 
Azndoode1 said:
The correct answer was actually +8.2...
So what's your point? It doesn't really help to get the correct answer if all you did was randomly add and subtract numbers until you found it. You need to understand what the variables and equations mean so you can get the correct answer without just guessing.
 
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