Confusion about the work done by an ideal gas

In summary, the conversation discusses the question of whether the work done by an expanding gas, in a piston system against a constant external pressure, is zero or not. The argument for work being zero is that the pressure inside the piston is zero, meaning there is no force pushing against the piston. However, this argument is incorrect because the pressure inside the piston would actually be equal to the external pressure, resulting in a non-quasistatic process. Additionally, the conversation touches on the concept of a massless piston and how it affects the calculation of work done by the gas. Ultimately, the conclusion is that the work done by the gas can be controlled by adjusting the external pressure on the piston.
  • #1
theblackfish
5
1
When an ideal gas,in a piston kind of system and whose equilibrium state is mentioned, is allowed to expand (piston is allowed to move and not gas leaking )against a constant external pressure very quickly, then, is the work done by gas zero or not zero ?

The argument for work being zero is that "the constituents take sometime to move towards the piston, as the piston moves away, and hence there's no pressure one the piston due to one side (inside ) and hence zero " I feel this argument is wrong because the pressure is 0 from the inside then the piston would be compressed by the external pressure.

Thank you in advance.
 
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  • #2
I think the question is if the expanding gas moves the piston (for example the piston ist locked the gas pressure is increased e.g. by isochoric heating and then the piston is unlocked) or if another force moves the piston and the gas is expanded by the increase of available room.

theblackfish said:
"the constituents take sometime to move towards the piston, as the piston moves away, and hence there's no pressure one the piston due to one side (inside ) and hence zero "

Wouldn't that mean that the piston moves faster than the gas is expanded? I think theoretically that effect could maybe play a role when approaching the speed of sound. So I think that you are correct with your last statement, but where do you get the quoted statement from?
 
  • #3
stockzahn said:
I think the question is if the expanding gas moves the piston (for example the piston ist locked the gas pressure is increased e.g. by isochoric heating and then the piston is unlocked) or if another force moves the piston and the gas is expanded by the increase of available room.
Wouldn't that mean that the piston moves faster than the gas is expanded? I think theoretically that effect could maybe play a role when approaching the speed of sound. So I think that you are correct with your last statement, but where do you get the quoted statement from?


Thats what my friends argued.
 
  • #4
1. If the external pressure is constant, and you want to move the piston quickly, then there must be an internal pressure which is greater than the external pressure. This gives rise to a non-quasistatic process, which blows the piston outwards. In general, it is not possible to calculate the work done by the gas.
2. Zero work is done if the gas expands into a vacuum. This is also a non-quasistatic process.
 
  • #5
When you release the piston, since the gas has inertia, a dilatation region begins to form in the gas immediately adjacent to the piston face. Within this region, assuming a massless piston, the pressure is virtually equal to the outside pressure (since the piston has no mass, it can accelerate at finite speed with even the slightest differential in pressure between inside and outside). As time progresses, the size of the dilatation region gets larger, and its leading edge travels backwards at the speed of sound within the gas (along the cylinder axis). So the phenomenon is similar to that of a sound wave.

In addition to this, there are viscous fluid stresses that develop within the gas that also contribute to the force per unit area exerted by the gas on the inside face of the piston. But, again, for a massless piston, Newton's third law tells us that the force exerted by the gas on the inside face of the piston must equal the constant external force exerted on the outside face of the piston.
 
  • #6
Chandra Prayaga said:
1. If the external pressure is constant, and you want to move the piston quickly, then there must be an internal pressure which is greater than the external pressure. This gives rise to a non-quasistatic process, which blows the piston outwards. In general, it is not possible to calculate the work done by the gas.
This is not correct. If the piston is massless, the work done by the gas is equal to the external pressure integrated over the change in volume. If the piston has mass, and before the piston comes to rest, the work done by the gas is equal to the kinetic energy of the piston plus the integral of the external pressure over the volume change (up to that time); if the piston has come to rest, so that the gas is at thermodynamic equilibrium in its final state, then the work done by the gas is equal to the integral of the external pressure over the volume change (up to the end of the process). This all follows from a straightforward force balance on the piston.

The bottom line is that, if you can control the external pressure exerted on the piston, then you are automatically controlling the work done by the gas on the piston.
 
Last edited:

Related to Confusion about the work done by an ideal gas

1. How is work done by an ideal gas defined?

The work done by an ideal gas is defined as the product of the force applied to the gas and the distance over which the force acts. This can also be expressed as the integral of the pressure with respect to the volume.

2. Is the work done by an ideal gas always positive?

No, the work done by an ideal gas can be positive, negative, or zero depending on the process being analyzed. If the gas expands, work is being done by the gas and the work done is positive. If the gas is compressed, work is being done on the gas and the work done is negative. If there is no change in volume, no work is being done and the work done is zero.

3. Can the work done by an ideal gas be negative?

Yes, the work done by an ideal gas can be negative if the gas is being compressed. This means that work is being done on the gas, rather than by the gas. The negative sign indicates that the direction of the force is opposite to the direction of the displacement.

4. How is the work done by an ideal gas related to its internal energy?

The work done by an ideal gas is related to its internal energy through the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. This means that the work done by an ideal gas can either increase or decrease its internal energy.

5. Does the work done by an ideal gas depend on the type of gas?

No, the work done by an ideal gas does not depend on the type of gas. It only depends on the pressure and volume of the gas, as well as the process being analyzed. This is because an ideal gas follows the ideal gas law, which states that the product of pressure and volume is constant for a given amount of gas at a constant temperature.

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