# Why do we use external pressure to find force exerted by gas on a piston in a closed vessel during its expansion?

I was reading a book in which i got confused due to this. A gas is in a vessel covered by a friction-less piston. Gas pressure is more than outer atmospheric pressure (which is constant) due to which it expands. Since it expands and move the piston outward, it does work on it which is calculated by force times displacement. In finding force we use Force = Pressure times Cross section area of piston. In finding this force to calculate work why are we using Pressure on piston by atmosphere and its weight and not pressure exerted by gas. Why? Can you also tell how the correct work done by gas can be calculated. Thanks.

For a frictionless, massless piston, the force per unit area that the gas exerts on the inner face of the piston is always equal to the force per unit area applied on the outer face of the piston (since the piston is massless, ma = 0). However, for an irreversible process, the force per unit area exerted by the gas at the interface is composed both of the "thermodynamic pressure" that one could determine from the equation of state for the gas (e.g., the ideal gas law) applied locally at the interface, plus "viscous stresses" in the gas that are determined by the rate at which the gas is deforming. Furthermore, if the expansion or compression is rapid enough, the conditions within the cylinder are not even uniform spatially. So figuring out the force per unit area that the gas exerts at the interface for irreversible expansion is problematic. However, fortunately, if the force per unit area on the outside is specified by manually imposing a force per unit area (e.g., if the atmosphere is imposing the pressure outside), the work done by the gas on the surroundings can still be determined. This is because the force per unit area exerted by the gas at the interface will adjust to always match the force per unit are on the outside (again, since the piston is massless and ma = 0).

In the case of a reversible process, the situation is much simpler, since here, viscous stresses are negligible, and the force that the gas exerts per unit area on the inner face of the piston can be determined by applying the ideal gas law (or real gas equation of state) to the spatially homogeneous gas in the cylinder. In this case, however, the force imposed on the outside must be controlled very carefully, so that it differs only slightly from the average inside gas pressure during the entire expansion or compression (and the gas pressure within the cylinder remains essentially uniform spatially).

why are we using Pressure on piston by atmosphere and its weight and not pressure exerted by gas.

Because the gas is expanding against the atmosphere. Put yourself in the shoes of the gas. Imagine that you're inside that container, trying to push the piston outwards. What are you pushing it against? The external pressure, right? The work done by the gas is, essentially, how much effort you are putting into pushing out the piston.

Can you also tell how the correct work done by gas can be calculated.

Since you're reading about work done, whatever book you are reading should really have a formula telling you that the work done by the gas is:

$$w = \int_1^2\!\mathrm{đ}w = \int_1^2\!p_{ex}\,\mathrm{d}V$$

where 1 and 2 are the initial and final states respectively.

• +1 except for that it is recommended to use a different đ symbol for inexact differentials. Commented Oct 3, 2015 at 18:17