# Work done by a gas

In the expression for work done by a gas, $$W=\int P \,\mathrm{d}V,$$ aren't we supposed to use internal pressure?

Moreover work done by gas is the work done by the force exerted by the gas, but everywhere I find people using external pressure instead of internal pressure.

• In equilibrium, the pressure exerted by a gas on the walls ("internal pressure") is equal to the pressure exerted by the walls on the gas ("external pressure"). Thermodynamics deals with transitions between a series of equilibrium states. – probably_someone Oct 23 '18 at 11:11
• @probably_someone The issue might be that if the pressure is changing, then it can't always be equal to the external pressure. – Aaron Stevens Oct 23 '18 at 11:12
• @probably_someone external pressure is equal to internal pressure if the piston is massless..but that is not true if the piston has sufficient mass – Alfred Mathew Oct 23 '18 at 11:15
• @probably_someone moreover in a thermodynamic process equilibrium need not be there in the intermediate stages – Alfred Mathew Oct 23 '18 at 11:17
• yes, we are assuming thermodynamic equilibrium, a quasi static change, a reversible process. And the piston is assumed to be massless, otherwise you are right, you need to consider the mass of the piston – Wolphram jonny Oct 23 '18 at 12:27

The work done by an expanding gas is the energy transferred to its surroundings. In effect, as the gas expands it is compressing its surroundings so the work done is the force exerted on the surroundings (i.e. the pressure of the surroundings times the area) times the distance moved.

The extreme case of this is a Joule expansion where a gas expands into a vacuum i.e. the pressure of the surroundings is zero. In this case the expanding gas does no work regardless of the initial pressure of the gas.

• force exerted on the surroundings is pressure exerted by the gas times the area,not external pressure – Alfred Mathew Oct 23 '18 at 11:19
• @AlfredMathew imagine you're compressing a spring with your hand. No matter how hard (how much force) you push with your hand, it's the $kx^2$ force from the spring that determines the work/energy that gets stored in it. – John Forkosh Oct 23 '18 at 11:23
• @John Forkosh That is because the spring is massless..and by newtons second law there must be no net force,hence the force we apply is equal to the force exerted by the spring – Alfred Mathew Oct 23 '18 at 11:35
• @JohnForkosh For the record, the force from the spring goes as $kx$. The energy stored goes as $kx^2$. – probably_someone Oct 23 '18 at 12:37

At the interface with the surroundings, by Newton's third law, the force per unit area exerted the gas on its surroundings is equal to the pressure of the surroundings on the gas. But, in an irreversible expansion or compression process, the pressure of the gas may not be uniform within the cylinder. So the pressures match only at the interface. In addition, viscous stresses contribute to the force per unit area exerted by the gas at the interface (as well as throughout the cylinder), so the equation of state (e.g., ideal gas law) cannot be used to establish the gas pressure within the cylinder or at the interface. The ideal gas law applies only if the gas is at thermodynamic equilibrium.

• your answer seems to be quite messed up..can you state it more clearly? – Alfred Mathew Oct 23 '18 at 12:50
• And my doubt was that in general shouldn't we use internal pressure in the expression for work even though in certain cases it maybe equal to external pressure – Alfred Mathew Oct 23 '18 at 12:52
• Sorry. I did the best I can. My main points were (1) that you can't use the ideal gas law to get the internal pressure for an irreversible process and (2) even so, at the interface with the surroundings, the pressure of the gas is equal to that of the surroundings. The pressure of the gas at the interface is always equal to that of the surroundings, irrespective of whether the process is reversible or irreversible. – Chester Miller Oct 23 '18 at 13:44

The work done by the gas on the piston is $$W_1 = \int P_{\text{int}} \, dV$$ where $$P_{\text{int}}$$ is the pressure of the gas right next to the piston. This is just a mild rephrasing of the definition of work. The work done by the piston on the outside is $$W_2 = \int P_{\text{ext}} \, dV$$ where $$P_{\text{ext}}$$ is the pressure of the external air right next to the piston. These two pressures may be different, so we may have $$W_1 \neq W_2$$.

For example, the two may differ if the piston has friction, with the difference $$W_1 - W_2$$ dissipated into heat. (Friction exists no matter how slowly the piston is moving, so this also holds for a quasistatic process.) Or the piston may be accelerating, in which case $$W_1 - W_2$$ goes into the piston's kinetic energy.

In high school physics, $$P_{\text{int}}$$ and $$P_{\text{ext}}$$ are always assumed to be the same, to keep things simple.

Work is done by the gas against the external pressure.If there is a case of free expansion of the gas (as in vacuum) the work done by the gas is zero as no opposing forces are present to prevent expansion of the gas, hence it is evident that work done by the gas is only due to external pressure. If the process is quasistatic(that is infinitesimally slow) then outside pressure is almost equal to internal pressure hence work done may be evaluated by considering the pressure of the gas, but if the process is not quasistatic then we must consider external pressure only.

• I dont understand how it is evident that work is done by external pressure only – Alfred Mathew Oct 23 '18 at 17:07