Understanding Heat Transfer in a Closed System with a Moving Piston

[vijayram]

Homework Statement

As shown in figure.

Homework Equations

dq=msdt[/B]

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.[/B]

 

[mfb]

There is no figure.

 

[vijayram]

There is no figure.

sir sorry for that,now i have edited it.

 

[Chestermiller]

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?

 

[vijayram]

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?

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?

 

[Chestermiller]

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?

All your answers are correct.

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).

 

[vijayram]

All your answers are correct.

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).

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.

 

[Chestermiller]

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.

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)

 

[vijayram]

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)

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?

 

[Chestermiller]

Does it mean the heat escaped from the cylinder moves to the rod and heats it?

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.

sir but,does how the piston is moved make any significance here?like heating the piston or physically moving it?

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?

 

[vijayram]

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?

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

 

[Chestermiller]

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

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

 

[vijayram]

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

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?

 

[Chestermiller]

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?

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.

 

[vijayram]

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.

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

 

[mfb]

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.

 

[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.

oh

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.

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?

 

[Chestermiller]

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

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.

 

[vijayram]

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.

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.

 

[mfb]

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.

 

[Chestermiller]

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.

Who says that the external force is constant.? I can apply whatever external force I desire manually, and the gas will have to respond. The force I apply does not have to be constant. And it can be over and above the atmospheric pressure (which, incidentally, is 100000 pascals, not 1 pascal).

 

[vijayram]

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.

Who says that the external force is constant.? I can apply whatever external force I desire manually, and the gas will have to respond. The force I apply does not have to be constant. And it can be over and above the atmospheric pressure (which, incidentally, is 100000 pascals, not 1 pascal).

yes sir understood it clearly,and yes it is 1lakh pascal,but what if the scenario is changed and the piston is heated by a heater?

 

[Chestermiller]

Do you mean the cylinder is heated by a heater? Are you saying that you can have heat transferred to the gas without a change in internal energy and without any work being done? Please provide a specific example of a problem which exemplifies what you are saying, and I can help you solve it.