Work done by the piston versus work done by the surrounding, but, the gas in the piston changes isothermally

Question

Another similar stack assumes the absence of the mass and the friction of the piston, but makes no specification regarding the change in temperature in the syringe. Here, we specify the temperature inside the syringe; the gas in the syringe is assumed to be isothermally.

The situations to be considered are as follows;

1 mol of ideal gas is isothermally expanded from $5.05 \times 10^5$ Pa to $1.01 \times 10^5$ Pa at 20 °C against a constant external pressure of $1.01 \times 10^5$ Pa.

My question is as follows;

• Is the above situation　physically possible ?
• If so, what controls could be used to achieve this change?

In this case,
the work that the gas in the syringe receives from the outside is; $$W_{outside \to inside} =- \int_{V_1}^{V_2} P_{out} dV = -P_{out}(V_2 - V_1) $$ On the other hand, if this can be regarded as an isothermal reversible process the work received by the outside from the gas inside the syringe is; $$W_{inside \to outside} =- \int_{V_1}^{V_2} P_{in} dV = -nRT\ln({V_2}/{V_1}) $$

This means that there must be a reasonable place for the energy equivalent to the difference between $W_{outside \to inside}$ and $W_{inside \to outside}$ to go.

Answer 1

The process you describe is not a reversible isothermal process because the external pressure is constant.

Your last equation only applies to a reversible isothermal ideal gas process. That requires the pressure of the gas to always be in equilibrium with the external pressure and that the external pressure continually decreases so that the product of the gas pressure and volume is always constant. That makes the gas temperature constant. Only then can you use the ideal gas law to calculate work.

Hope this helps.

Answer 2

In the above case, work done by system = $-P_{out}\Delta V=-P(V_2-V_1)$.
As $V_2>V_1$ so, $-P(V_2-V_1)<0$, hence work is done by the system.
Now the point is why can't we use the formula $-\int_{V_1}^{V_2} P_{in} dV = -nRT\ln({V_2}/{V_1})$, this we have to figure out.
We can see that system's internal pressure is very large than surrounding's pressure, so there is large unbalanced force on the surface of piston towards surroundings causing the system to expand very fast. In the intermediate steps of expansion, we can't define the state variables like pressure and temperature of the gas. Because it takes time for these variables to attain an equilibrium value. So the system expands fastly, till its pressure matches the surrounding pressure. After reaching that pressure value the temperature and pressure gradients created in the system get a uniform value.
So, $P_{out}(V_2-V_1)$ amount of energy flows from system to surroundings.
So, Work done by system=-Work done on system.
One can see that the convention of work used here in thermodynamics is somewhat opposite to that used in mechanics. The convention which we have used above is always with respect to system, while in mechanics it is wih resect to the agent which does work. Here positive work done means work is done on system (energy flows into system), while in mechanics if an agent does positive work on a body, then equivalently the sign attributed to the work done by body is negative and it means energy flows into the body.
Hope it is clear to you now!