# Dilemma on work done by gas in isothermal contraction

Consider an ideal gas inside a cylinder undergoing reversible isothermal contraction. Now if we consider the forces acting on the gas inside the cylinder the force acting is $P_{ext}\cdot A$, however the gas inside the cylinder will also exert a force on the external gas equal to $P_{int}\cdot A$. Now as per Newton's third law there should be an equal and opposite force acting on the gas inside the cylinder which is equal to $P_{int}\cdot A$, now since it is reversible $P_{int}=P_{ext}$, therefore total force acting on the gas inside the cylinder is $2P_{ext}$. But when we consider the work done on the gas we take only $P_{ext}$ and not $2P_{ext}$, why is that?

• you are looking the wrong the force which internal gas exert $P_{int}.A$ on the external gas is only the reaction of external force $P_{ext}.A$ – 12sa Dec 30 '15 at 11:48

In thermodynamics, you always consider the work performed by one system on another. In your example, the external gas performs work on the internal gas (and of course you can always change your perspective so that the internal gas does the opposite of this work on the external gas). In considering the work performed by the external gas on the internal gas, the relevant quantities are $P_{ext}$ together with any losses to friction. Assuming constant $P_{ext}$, if the piston is moved a distance $\Delta x$ then the external gas will perform work $P_{ext}A\Delta x$ on the piston, and (depending on friction) some fraction of the energy will be transferred through the piston to system 2. Note also that the net work performed on the piston between the two systems is zero in the [isostatic]&&[reversible] limit (i.e. where $P_{ext}\approx P_{int}$).
1. When $P_{ext}\neq P_{int}$ (i.e. the expansion is irreversible), where does the extra energy go?