# Work done during expansion of a gas at constant pressure

Discussion in 'Physics' started by logearav, Nov 14, 2011.

1. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Revered members,
Consider this scenario. A cylinder is fitted with a frictionless and free to move piston of area of cross-section A. Let a gas of volume V1 be enclosed in the cylinder. Now if we heat this enclosed gas its volume increases and the due to heating the piston moves up which is an indicator for decrease in pressure. But my book says, increase in volume due to temperature rise will take place at constant pressure equal to atmospheric pressure. I cant understand. How the pressure will remain constant, when the piston moves up?

2. ### davebee Well-Known Member

Oct 22, 2008
539
46
This looks like an application of the ideal gas law PV=NRT:

pressure times volume = (number of molecules) times (constant R) times temperature

First, I asssume the piston must be horizontal to eliminate the effect of gravity, so it would move "out" not "up".

Second, the piston moving "up" would not necessarily indicate a decrease in pressure, would it?

The pressure remains constant because with a frictionless piston, any pressure difference would result in the piston moving until there is no pressure difference.

As you can see from the ideal gas law, if temperature increases, the volume will increase proportionally, if every other parameter remains the same.

3. ### BillO Distinguished Member

Nov 24, 2008
985
136
logearav,

You have an unerring ability to get yourself confused. I am not sure where it comes from, but you certainly seem to get distracted by things that are not material to the question at hand.

As stated, and I am almost certain this must have be mentioned in your texts, Boyle's ideal gas law, as previously stated, is what governs the relationship between volume, temperature and pressure.

Given; PV=nRT

We can see that if P is a constant, then V is proportional to T. This means that at a constant pressure, as temperature (T) increases, so must the volume (V).

The two books you talk about must be mentioning different applications for Boyle's law. You must read, and try to understand what the conditions are for the discussion.

PV=nRT is the defining equation. Let it be your guide. If you are still confused, scan the pertinent sections of each book and post it here. we can then have a look.

4. ### logearav Thread Starter Member

Aug 19, 2011
248
0

Thanks for the replies davebee and BillO.
@ davebee,
Sorry, I am a novice. Just to clarify i am asking
When the piston moves down, does it not compress the molecules, which means increase in pressure? Also, when it moves up, there is no compression so pressure decreases. If temperature rises, gas molecules expand which in turn move the piston upwards. So the pressure decreases.
P ∞ 1/V

5. ### BillO Distinguished Member

Nov 24, 2008
985
136
According to PV=nRT

If (P)ressure stays the same, and (T)emperature goes down, then the (V)olume must also go down. In this case the piston will move in.

Also,

If Pressure stays the same, and Temperature goes up, then the Volume must also go up. In this case the piston will move out.

Last edited: Nov 18, 2011
logearav likes this.
6. ### davebee Well-Known Member

Oct 22, 2008
539
46
The intent of this experiment is that the only forces acting on the piston are the gas pressures on the two sides of the piston.

"When the piston moves down, does it not compress the molecules"
"when it moves up, there is no compression so pressure decreases"

logearav, what forces move the piston like that?

The answer is that there is no force moving the piston like that, so no compression or decompression takes place.

You need to apply an external force to compress or decompress the gas, but no external force is applied, so there is no compression or decompression of the gas. The gas remains at constant pressure.

logearav likes this.
7. ### logearav Thread Starter Member

Aug 19, 2011
248
0
@davebee,
So, when we heat the gas, volume of the gas molecules increases, and when we decrease the temperature or cool the gas, volume will decrease. Piston wont move out or in, when heating or cooling is done. Am I right?

8. ### davebee Well-Known Member

Oct 22, 2008
539
46
yes, you're right - when no change in temperature takes place, the piston does not move.

9. ### logearav Thread Starter Member

Aug 19, 2011
248
0
But when we heat or cool the gas, there is change in temperature. So, how piston will remain stationary?

10. ### BillO Distinguished Member

Nov 24, 2008
985
136
If you heat or cool the gas, the piston will not remain stationary.

I updated post #5 to include the piston motion in each case. Look back to that post.