# ¿How does pressure decrease in an isothermal process when heat is transfered?

Let's suppose there is a gas(ideal) inside a cilinder with a frictionless, massless piston. The gas is under a pressure $P_{1}$ and has volume $V_{1}$ at a temperature $T$. We carry out a isothermal process (I have no idea how to conduct one, but i'm assuming I do). Here's a picture. Heat is transfered into the system.

Here's my question. What I understand is that for the piston to start moving, more force is required to defeat the initial pressure. Therefore, the pressure increases so does the volume. I kwow I'm wrong because according to the state equation $PV=RTN$ pressure should decrease. I hope you could provide me a deepest insight about what is happening at the molecular level to understand this process.

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You should probably make your second question a separate question, as it's pretty unrelated to your first. – Jahan Claes Nov 13 '15 at 4:02
Thanks. I will delete that part. – Omar Nov 13 '15 at 4:02

## 1 Answer

In the picture you've just drawn, pressure is constant for the entire process. Why? Well, the forces on the piston are $PA$ and $-mg$, and they have to sum to zero for the piston not to accelerate off either up or down. $A$, $m$, and $g$ don't change, so $P$ doesn't change either. Then the only way for V to change is for T to increase. So you haven't drawn an isothermal process.

But let's pretend you did draw an isothermal process. Then $T$ is constant, so either $P$ decreases and $V$ increases or vice-versa. Let's consider what has to happen to increase $V$, decrease $P$, and keep $T$ constant.

First: if we're going to increase $V$, the gas is going to do work on the environment. So, we need to supply some heat $Q$ which is exactly equal to the work done. So we're going to heat this container during this process, and carefully control the heat to keep $T$ constant. Alternatively, we're going to perform this process VERY SLOWLY, and allow the gas time to gain heat from the environment. Second, we need the gas to expand. How do we do that? We have to decrease the external force on the piston. We either take some weight off the top of the piston (decrease m in your picture), or grab the piston and pull up. In general, you're correct in saying that the pressure of $P$ is not going to make the gas expand without us doing something.

What I'm trying to get across is that these processes don't happen spontaneously. The ideal gas law has a lot of constants that can vary in a lot of different ways; if you want a specific process to happen (e.g. adiabatic, isothermal, isobaric, etc) you have to do something VERY SPECIFIC to the gas. In this case, you need to simultaneously provide heat AND pull on the the piston to create an isothermal process. You asked why the container would expand if $P$ was decreasing; the answer is, $P$ decreased because YOU did something to expand the container.

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First, thank you for your very detalied answer. So what I used to think was that at constant temperature, you just need to add heat into the system and the gas would expand pushing the piston. But I was wrong, so I just imagined this piston is formed of small pieces so that We can take them off very slowly. That way, the pressure will decrease allowing the gas to expand. I hope this is what you meant. However, even without heat getting into the system, If we pull on the piston, the gas would expand anyway or am I mistaken? – Omar Nov 13 '15 at 4:50
@Omar Without heat going into the system, if you pull on the piston, the pressure will decrease and the gas will expand, and the temperature will decrease. This last part will make it non-isothermal. – Jahan Claes Nov 13 '15 at 4:58
So there's no way to control temperature wihout introducing heat into the system. I wonder why. However, with your answer I think I'm getting closer to get it. I'll keep researching more. Thank you :) – Omar Nov 13 '15 at 5:05
@Omar If you don't let heat into the system, the gas will do work as the piston expands, and so the gas will lose energy. – Jahan Claes Nov 14 '15 at 5:02