External Work done by a tank containing air

[Idris Sayyad]

Homework Statement

A tank containing air is stirred by a paddle wheel. The work input to the paddle wheel is 9000 kj and the heat transferred to the surroundings from the tank is 3000 kj. The external work done by the system is ?

Homework Equations

dU = Q-W

The Attempt at a Solution

Work input = -9000 kj
heat given to the surroundings = -3000 kj

The answer given is Zero, but no explanation has been given .
Is it zero because there is no movement of the boundary in this case?

 

[Andrew Mason]

Homework Statement

A tank containing air is stirred by a paddle wheel. The work input to the paddle wheel is 9000 kj and the heat transferred to the surroundings from the tank is 3000 kj. The external work done by the system is ?

Homework Equations

dU = Q-W

The Attempt at a Solution

Work input = -9000 kj
heat given to the surroundings = -3000 kj

The answer given is Zero, but no explanation has been given .
Is it zero because there is no movement of the boundary in this case?

Welcome to PF Idris Sayed!

The air is in a tank, which I assume is a fixed volume. Does the volume of air change? If not, how can the system do work?

AM

 

[Idris Sayyad]

Thank you sir :)

There is no mention of the change of volume of air in the question. So if the answer is Zero, then volume is fixed has to be assumed I guess in this case.

 

[Andrew Mason]

Thank you sir :)

There is no mention of the change of volume of air in the question. So if the answer is Zero, then volume is fixed has to be assumed I guess in this case.

I think it is the other way around. If the air is in a tank, its volume does not change. External work can only be done if there is a change in volume (i.e. a pressure applied over a change in volume = force x distance). Since there is no change in volume, no external work is done by the air.

AM

 

[Idris Sayyad]

I think it is the other way around. If the air is in a tank, its volume does not change. External work can only be done if there is a change in volume (i.e. a pressure applied over a change in volume = force x distance). Since there is no change in volume, no external work is done by the air.

AM

Thank you sir , This cleared my doubt :)