Kinetic Theory - Mass on a piston

[danielbaker453]

Homework Statement

A cylinder containing an ideal gas is in vertical position and has a piston of mass $M$ that is able to move up or down without friction (Fig. 13.2). If the temperature is increases,

...Fig (13.2)
(a) both $p$ and $V$ of the gas will change.
(b) only $p$ will increase according to Charles' Law.
(c) $V$ will change but not $p$.
(d) $p$ will change but not $V$.

Homework Equations

$pV=nRT$
$ΔU=q+w$[/B]

The Attempt at a Solution

The answer that is given is (c) and the reasoning is that the pressure (=Mg/A) is constant. But I don't get it. I think that the pressure of the gas will increase because as the temperature increases, the molecules will start hitting the walls of the container with greater speed causing an increase in pressure. Also as the temperature increases, the internal energy will increase and work will be done by the system on the piston thus lifting it up. This will cause an increase in volume. So, as the pressure and the volume are changing the correct answer would be (a).
Could someone please explain to me why I'm wrong?

 

[CWatters]

The answer that is given is (c) and the reasoning is that the pressure (=Mg/A) is constant. But I don't get it. I think that the pressure of the gas will increase because as the temperature increases, the molecules will start hitting the walls of the container with greater speed causing an increase in pressure.

That's correct but what happens next is incorrect. I mean what happens if the pressure inside is greater than Mg/A?

 

[Chestermiller]

As the volume increases, the molecules get further apart, so the frequency of collisions with the piston decreases.

 

[danielbaker453]

That's correct but what happens next is incorrect. I mean what happens if the pressure inside is greater than Mg/A?

If the pressure inside is greater, the piston will be pushed up. This means that the pressure is changing and so is the volume(due to expansion). Am I right?

 

[Chestermiller]

If the pressure inside is greater, the piston will be pushed up. This means that the pressure is changing and so is the volume(due to expansion). Am I right?

No. The pressure will stay constant.

 

[danielbaker453]

No. The pressure will stay constant.

But the pressure exerted by the gas has increased. I do understand that the pressure on the gas due to the mass 'M' is constant (=Mg/A) but isn't 'p' the pressure exerted by the gas?

 

[Chestermiller]

But the pressure exerted by the gas has increased. I do understand that the pressure on the gas due to the mass 'M' is constant (=Mg/A) but isn't 'p' the pressure exerted by the gas?

If the piston is massless and frictionless, what does the force balance on the piston tell you?

 

[danielbaker453]

If the piston is massless and frictionless, what does the force balance on the piston tell you?

In that case, the only force on the piston is that exerted by the gas molecules.

 

[Chestermiller]

In that case, the only force on the piston is that exerted by the gas molecules.

Sorry. I forgot that the piston is supposed to have mass. So, let me rephrase my question: If the piston is frictionless, what does the force balance on the piston tell you?

 

[danielbaker453]

Sorry. I forgot that the piston is supposed to have mass. So, let me rephrase my question: If the piston is frictionless, what does the force balance on the piston tell you?

Got it!
Now, the force exerted by the gas equals the weight of the mass 'M'.
F_gas=Mg
If we divide both sides by the area 'A'
P_gas=Mg/A
The pressure will be constant.
Thank you very much for your help!
P.S. since here P_external=P_internal, is this a case of reversible expansion?

 

[Chestermiller]

Got it!
Now, the force exerted by the gas equals the weight of the mass 'M'.
F_gas=Mg
If we divide both sides by the area 'A'
P_gas=Mg/A
The pressure will be constant.
Thank you very much for your help!
P.S. since here P_external=P_internal, is this a case of reversible expansion?

The gas force per unit area on the inside face of the piston always matches the $P_{ext}$ (by Newton's 3rd law), irrespective of whether the expansion is reversible. But, in an irreversible expansion, the gas pressure in the cylinder varies with spatial position and the force per unit area at the piston face includes viscous stresses. So, in an irreversible expansion, the ideal gas law (or other equation of state) cannot be applied globally to the gas.

 

[CWatters]

+1

At some point the piston will stop accelerating. If it's not accelerating the net force on it is zero so the force due to the pressure of the gas must equal the force due to gravity.