# A doubt in thermodynamics

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1. Oct 28, 2016

### vijayram

1. The problem statement, all variables and given/known data
As shown in figure.

2. Relevant equations
dq=msdt

3. The attempt at a solution
I am unable to understand the question,what is the significance of the moving piston if the ends of the conducting rod is maintained at constant temperature.It is enough if someone could explain the question and the logic.

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2. Oct 28, 2016

### Staff: Mentor

There is no figure.

3. Oct 28, 2016

### vijayram

sir sorry for that,now i have edited it.

4. Oct 28, 2016

### Staff: Mentor

This is a problem in applying the First Law of Thermodynamics, not the Second Law.

1. If the rate of heat flow through the rod is constant, then the gas temperature is (a) constant or (b) not constant?

2. From the answer to question 1, the internal energy U of the gas is (a) constant or (b) not constant?

3. From the answer to question 2, the rate of doing work on the gas is (a) equal to the rate of heat flow through the rod or (b) not equal to the rate of heat flow through the rod?

5. Oct 28, 2016

### vijayram

thank you for replying sir,but isn't it already given the temperature of the gas is constant 300k.Anyway the internal energy of the gas is constant but how is work done on the gas equal to the rate of heat flow through the rod?
From first law of thermodynamics we can say that.
dq=du+dw.
since du=0
the rate of heat of heat flow into the gas is equal to the work done on the gas,but how is related to the rod?

6. Oct 28, 2016

### Staff: Mentor

Now,
the rod is removing heat from the gas. The rate of heat flow through the rod is equal to the rate of heat removed from the gas. Or, equivalently, the rate of heat flow into the gas is equal to minus the rate of heat flow through the rod (i.e., $\dot{Q}$ is negative).

7. Oct 28, 2016

### vijayram

sir,why should the rod remove heat from the gas?we don't know how the piston is moving,we can say,piston is moving because of some external heat or if piston is moved physically,heat can escape out of the system.

8. Oct 28, 2016

### Staff: Mentor

What do the words "Neglect any kind of heat loss from the system (other than through the rod)" mean to you? (Italics are mine for clarification)

9. Oct 28, 2016

### vijayram

Does it mean the heat escaped from the cylinder moves to the rod and heats it?i thought about this but i couldn't be sure?sir but,does how the piston is moved make any significance here?like heating the piston or physically moving it?

10. Oct 28, 2016

### Staff: Mentor

It flows through the rod at a constant rate from the gas to the cold reservoir. The rod temperature profile does not change with time.
The implication of the problem statement is that the piston is being physically moved.

The person who devised this problem statement thought that he was stating it clearly enough. But, apparently, it was not clear enough for you. So I will try to state it more clearly:

Assume that no heat enters or leaves the gas except through the rod. Work is being done by applying external force to the piston. Assume that the work is done reversibly.

Now under these assumptions, please solve the problem and see if you answer matches any of those given?

11. Oct 28, 2016

### vijayram

Thank you very much sir but if the piston is moved by a heater,how would the scenario change?

12. Oct 28, 2016

### Staff: Mentor

Why don't you solve the problem as stated already? I did, and I obtained one of the choices they offered.

13. Oct 28, 2016

### vijayram

sir i solved the problem and got k/100R as the answer,but i have one more additional doubt,the piston is massless so net force on the piston should be zero or external pressure should be equal to internal pressure,or internal pressure should be equal to atmospheric pressure,but the pressure obtained using gas laws is different as temperature is constant.So isn't this a contradiction,how to explain it sir?

14. Oct 28, 2016

### Staff: Mentor

External pressure is slightly higher than internal pressure for a reversible compression, and the ideal gas law can be applied even if the temperature is constant. So I don't see any contradictions. The pressure is changing during the compression.

15. Oct 28, 2016

### vijayram

but the piston is massless so according to newton's law f=ma,m=0 so net external force should be equal to zero.

16. Oct 28, 2016

### Staff: Mentor

The external force on the piston is equal to the internal force on the piston. We do not know where the external force on the piston comes from (surrounding air, some machine, ...), but it does not matter.

17. Oct 28, 2016

### vijayram

oh
yes i think the same sir,so atmospheric pressure is not equal to internal pressure of the gas and it could be found only through the gas laws.Am i right sir?

18. Oct 28, 2016

### Staff: Mentor

The net force is essentially zero. The external pressure essentially matches the gas pressure for a reversible compression. As long as the piston moves, however slowly, work is being done on the gas. It may take a long time, however.

19. Oct 28, 2016

### vijayram

but sir,external pressure is constant and is equal to 1 pascal but internal pressure changes with the position of the piston.And the numbers didn't match sir.Could it be like this sir?when the external agent is also taken into account,it comphensates,and the piston cannot move on it's own.

20. Oct 28, 2016

### Staff: Mentor

External pressure does not matter.

You are confusing yourself needlessly by adding things to the problem that are not there.

Something moves the piston, that is all you have to consider here.