Work done on a gas enclosed in a container

According to everywhere work done on system of a gas is equal to $$p\,\mathrm dV$$ but here $$p$$ is just the external pressure. Shouldn't internal forces be considered while calculating work done thus giving a place for internal pressure as well? Shouldn't the $$\mathrm dW$$ be equal to $$(p_1 -p_2)\,\mathrm dV$$ where 1 is internal and 2 is external?

• Why do you think $p$ is the external pressure? – flaudemus Feb 19 at 15:45
• Like when it is done in the case of irreversible process and in case of reversible process p int=p ext – dawood mansoor Feb 19 at 15:48
• Why did someone down vote this please clarify – dawood mansoor Feb 19 at 15:50
• For all cases, at the boundary between the system and surroundings, $p_{int}=p_{ext}$. However, for an irreversible process, the value of $p_{int}$ at the boundary is not given by the ideal gas law. The ideal gas law only gives the correct value for reversible processes (close to thermodynamic equilibrium). – Chet Miller Feb 19 at 15:58
• I know but why we don't take the internal force work done which is exerted by internnal pressure – dawood mansoor Feb 19 at 16:14

I want to explain this usage, i.e. why use internal pressure, by using a container with a piston as an example. You can make similar thinking for other and more general case. If we take a look of the piston, it has 3 forces on it: external pressure, internal pressure and frictional force on its perimeter interface with the container wall. For most problems, we assume there is no friction and the piston moves very slowly. That means frictional force is zero and acceleration is negligible.

A free body diagram can tell us that the external pressure equals the internal pressure.

Work is the multiplication of force and piston travel distance. Thus, the external pressure produces the same amount of work as that from the internal pressure but with opposite sign.

Now, in a more generalized case, where friction cannot be ignored, the work done by the external force would not be the same as the work done by the internal pressure; there is an extra work done by the frictional force.

By whatever way, work done by the internal pressure is completely used for or by the gas system whereas work done by external pressure involves extra work such as that from frictional force and cannot be directly used for gas system calculation.

If it is not a piston container system but a gas gan, when you squeeze the can by using very high external pressure, the work done by the external pressure, in this case, includes the can plastic deformation energy.

To be simple and less prone to make mistake, textbook tells people to use internal pressure for gas system calculation. If you want and you are very clear in mind on the force involved in your system, you can use external pressure for work calculation. Just be careful about the extra work and the sign of the work.

Note $$(p_{external}-p_{internal}) \times dA$$ is the force on the piston. $$(p_{external}-p_{internal}) \times dV$$ is meaningless as the external pressure is not directly applied on the gas.

• Got it thanks. I had to tak3 the gas as system but I was taking gas and piston as system – dawood mansoor Feb 19 at 18:24

First of all, if you are talking about the ideal compression where $$P_{ext}$$ is just bigger than $$P_{int}$$ by $$dP$$.

Secondly, when we learnt Newton's laws, we do agree that when we talk about forces acting on a body, we do not take the forces acting by the body on other things. The pressure $$P$$ applied externally is intended to cause a change in volume of the gas.

There won't be sense to actually talk about internal pressure involved in cancelling this effect off. $$W=\int PdV$$ is in fact, just a re-written form of: $$W=\int Fdx$$ $$=\int PAdx$$ $$=\int PdV$$

To make it clear. This is no isothermal or adiabatic compression. This is only a constant pressure use. True, that the internal pressure will change, but we always assume that the external pressure is just equal to the internal pressure.

• That's Newton's law while considering acceleration of a system but in the case of applying work energy theorem we take work done by internal as well as work done by external – dawood mansoor Feb 19 at 16:23
• The gas does not lose energy in this process at all. It gains energy. And only due to an external force. – KV18 Feb 19 at 16:29
• i think there is communication gapp here. I am asking why the work done by internal force is not counted here?? – dawood mansoor Feb 19 at 17:51
• When you use the first principle of thermodynamics, only external work has to be taken into account. It is complety different from the work energy theorem in which you take into account internal and external work. – Vincent Fraticelli Feb 19 at 19:49

At the interface between the system (say, the gas) and the surroundings (say, the inside face of the piston) where work is being done, the "internal pressure" of the system is always equal to the "external pressure" of the surroundings. And, in a reversible process, the internal pressure can be determined by the ideal gas law (or other equation of state, say, for a real gas).

However, in an irreversible process, the gas is not at thermodynamic equilibrium throughout the process. And, since the ideal gas law (or other equation of state) is correct only at thermodynamic equilibrium, the internal pressure of the gas at the boundary where work is being done cannot be correctly calculated using the ideal gas law. This is because viscous (dissipative) stresses also contribute to the internal pressure at the interface in the case of an irreversible process. As a result, in irreversible processes, the internal pressure depends not just on the gas volume and temperature but also on the time rate of change of gas volume. So the ideal gas law can't be used. For an irreversible process, we can determine the work done only by controlling the interface pressure externally, say by using an automatic control system or by removing specified weight from the piston.