# Isobaric Process

I was studying the thermodynamic processes, particularly the isobaric process where the pressure remains constant. While solving a particular question (which I couldn't) I saw that for a gas enclosed in a vessel with a movable piston the pressure is said to be constant while the gas expands due to heat given to the system (assuming no loss of heat to the surrounding). I just couldn't understand how the pressure can be constant because if we give heat to the system it will increase the kinetic energy of the molecules and thus increase the pressure which will lead to expansion, whereas in the solution what I find is that the pressure is constant because of the fact that only atmospheric pressure is acting on the system. My ultimate doubt is what pressure are we talking about, the pressure exerted by the gas or the pressure by the external agency (surroundings)?

• Net pressure, pressure due to kinetic energy of the molecules and atmospheric pressure.(on piston)....... Dec 7, 2019 at 4:51
• Does this answer your question? Definition of isobaric process Dec 7, 2019 at 4:58

$$dW = P_{external} dV$$
Work done on a gas is dependent only on the external pressure applied to the gas, which is constant (equal to $$P_{atmosphere}$$ ) in most cases. It is only in quasistatic processes that $$P_{external} = P_{gas}$$ at every step of the process.
Unless it is mentioned take $$\Delta W = P_{atmosphere} \Delta V$$ to find out the work done.
For a massless, frictionless piston, according to Newton's 2nd law, the external force per unit area exerted by the gas on the piston is always equal to the force per unit area $$P_{ext}$$ exerted by the surroundings on the piston. If the deformation is quasi-static and reversible, the force per unit area of the gas is also equal to the gas pressure calculated by the ideal gas law. However, for a rapid (non-quasistaitc) expansion or compression, the ideal gas law cannot be used to predict the force per unit area because the ideal gas law applies only to equilibrium (quasi-static) situations. But, if the external force per unit area is known, this can always be used to calculate the work for such cases.