# Why would the gas do work when the piston is pulled by external force

Taking an adiabatic process as example , from the first law of thermodynamics $$\Delta U=-W... (1)$$ .Now if I pull up the piston applying force from my side , the work is done by me on the system so the sign is negative, so from (1) , $$\Delta U$$ is positive , but I know that temperature should decrease and so should internal energy , then is it that I am taking the wrong signs
moreover , why would the gas lose its internal energy anyway , after all , its the external force applied by me which is moving the piston , so why should the gas inside do any work and lose temperature ?

• Have a look at the signs convention you're using. – FGSUZ Feb 8 at 16:53
• Let’s say the system is initially in equilibrium. If you slowly pull on the piston the pressure of the gas drop, volume increases and temperature drops. You then stop pulling and maintain the force to hold the piston in place. You have done positive work on behalf of the system so the internal energy of the gas has decreased. But if you now slowly allow the piston to return to its original state (so that you no longer have to apply a force to keep it in place) you will have done negative work increasing the internal energy. – Bob D Feb 8 at 17:53

Imagine experiencing in a vacuum to avoid the issue of atmospheric pressure outside the piston. A wedge prevents the piston from mounting and is removed at the beginning of the experiment.

If you pull the piston very quickly, faster than the speed of sound, there will be no resistance: the outside work will be zero. It is a Joule expansion, without work and $$\Delta U=0$$.

If you let the piston go up slowly, it is the gas that pushes the piston and you must exert a force directed downwards which prevents it from accelerating. In this case $$-W<0$$, and therefore $$\Delta U<0$$ because the gas provides work.

Sorry for my english.

If the gas in the cylinder is considered the “system” then you are the “ surroundings”. Initially the gas and you are in equilibrium with no force applied to one another.

From the perspective of the system (gas) when you pull on the piston you are reducing the external pressure allowing it to do positive expansion work on the surroundings (you) losing internal energy while the surroundings (you) gains internal energy. On the other hand from the perspective of the surroundings (you) you are the one doing positive work on the system and losing internal energy while the system should be gaining internal energy.

Now let’s reverse the process. You reduce your force and since the gas pressure is lower than your pressure you feel the gas is pulling on you. From your perspective the gas is doing work on you increasing your internal energy while reducing its internal energy. But from the gas perspective you are doing compression work on it increasing its internal energy while you are losing internal energy, until eventually, if the processes are carried out very slowly (reversibly), the gas and you return to your original states with no net work done and no change in internal energy

Bottom line: how you perceive the work done depends on whether you look at it from the systems perspective or surroundings perspective.

Hope this helps

Think about it like this, q in adiabatic process is zero. So the only way to transfer energy is by doing work or extracting work. Now when a gas expands it has to do work against the external pressure to increase its volume which comes at the expense of its own internal energy, hence internal energy decreases.