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I came across $ΔU=Q+W_o $ where $W_o$ represents the work done on the system. I also came across the formula $ΔU=Q-W_B$ where $W_B$ represents the work done by the system (gas). My question is, since the work done by the system is equal to the negative of the work done on the system, is the magnitude of the force exerted by the system on the piston and vice versa assumed to be the same?

I have also learnt that according to the work energy theorem, net work done on a system= change in it's total kinetic energy, thus, if we do 30J of work on the system, the kinetic energy should increase by 30J, and since the total internal energy of an ideal gas is equal to the kinetic energy of the molecules, the internal energy of the gas increases by 30J.

Is my logic correct?

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  • $\begingroup$ Thats all correct. Occasionally we can also change the height of some mass in the system and that is a type of potential energy (because we could get work out of it by lowering it back down). $F=mg$ and is constant, so work = $\Delta PE = mgh$. If stuff flows up or down for example $\endgroup$
    – Al Brown
    Aug 4 at 3:12
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About your second answer, Yes your logic is right.

But if the temperature (temperature, not internal energy) of the surroundings is lesser than that of the system(after adding 30J), some of the internal energy of the system will be transferred, from the system, to the surroundings.

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First question: Yes, that's Newton's third law!

Second question: Yes, assuming that heat is not also transferred. If $Q=0$, then $\Delta U=W_{0}$.

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  • $\begingroup$ Short and sweet! Thanks a bunch!!! $\endgroup$
    – Arun
    Jan 23 '20 at 2:04
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My question is, since the work done by the system is equal to the negative of the work done on the system, is the magnitude of the force exerted by the system on the piston and vice versa assumed to be the same?

The magnitude of the work is the same. The first equation is commonly used in chemistry. The second equation is the more mainstream used in physics and engineering. It doesn't matter which is used as long as you are consistent. Consistent in this regard is understanding that when work is done by the system (e.g., expansion work by a gas), it takes internal energy away from the system, and when work is done on the system (e.g., compression work on a gas) it adds to the internal energy of the system.

I have also learnt that according to the work energy theorem, net work done on a system= change in it's total kinetic energy, thus, if we do 30J of work on the system, the kinetic energy should increase by 30J, and since the total internal energy of an ideal gas is equal to the kinetic energy of the molecules, the internal energy of the gas increases by 30J.

The work energy theorem generally applies to the velocity and kinetic energy of macroscopic bodies. For a gas in a cylinder fitted with a piston, that would be the kinetic energy associated with the center of mass of the gas as a whole with respect to an external frame of reference. Visualize the cylinder of gas as a whole moving with some velocity. The principle is not applicable to the internal microscopic kinetic energy of the gas.

That being said, your logic that 30 J pf work done on an ideal gas increases its internal kinetic energy by 30 J is correct, provided that there is no heat transfer out of the system during the compression. In other words your process needs to be adiabatic so that $Q=0$. And that's because the first law allows for the possibility of energy transfer being in the form of heat and/or work per the equation.

$$\Delta U=Q-W$$

UPDATE:

This will respond to your follow up questions.

Is the reason why the work done by the system and piston are equal in magnitude but opposite in sign, as mentioned by Shura Zeryck(another contributor), because of Newton's third law of motion?

If you asking if the reason the signs for work in the two equations for the First Law law are different is because of Newton's third law, it is not. As explained above the difference in signs is simply due to different sign conventions. When work done by a system on the surroundings and $Q=0$, the result is a decrease in internal energy. In the first equation the convention is to call this negative work. That makes $\Delta U$ negative. In the second equation the convention is to call this positive work. That also results in $\Delta U$ being negative. In either case the magnitude of the work is the same and the two versions ares consistent. Beyond that, there is no other reason for the difference.

If you are asking if Newton's third law applies to work in general, then of course it applies to the forces that are doing work. But in terms of the net work done on each of the interacting objects is concerned, one applies Newton's second law.

Another question I have is that generally, when positive work is done on the system, we state that "work is done on the system" while when negative work is done on the system, we state that "work is being done by the system" My question is why is it so? Do we conventionally define work from the frame in which work done is positive? ?

I'll answer the second question first. In thermodynamics we conventionally define work in the reference frame of the "system". Everything outside the system is lumped together and called the "surroundings". The focus is on the system and its properties. The properties of the surroundings are not generally specified. Normally only the work or heat transferred to or from the surroundings, and its entropy change, is of concern.

As to the first question, just to be clear, when you say "when positive work is done on the system, we state that work is done on the system while when negative work is done on the system, we state that work is being done by the system" be aware you are only referring to the version of the first law $\Delta U=Q+W$. The second version $\Delta U=Q-W$ is more commonly used. For the second version, just reverse the terms "positive" and "negative".

That being said, the reason why we specify work done by or on the system is to assign the proper signs for the values of $W$ in the applicable first law equation, as I previously discussed.

Hope this helps.

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  • $\begingroup$ Thanks for the answer! Is the reason why the work done by the system and piston are equal in magnitude but opposite in sign, as mentioned by Shura Zeryck(another contributor), because of Newton's third law of motion? Another question I have is that generally, when positive work is done on the system, we state that "work is done on the system" while when negative work is done on the system, we state that "work is being done by the system" My question is why is it so? Do we conventionally define work from the frame in which work done is positive? $\endgroup$
    – Arun
    Jan 23 '20 at 2:03
  • $\begingroup$ @Arun I will be updating my answer to respond to these follow up questions. Please stand by $\endgroup$
    – Bob D
    Jan 23 '20 at 10:20
  • $\begingroup$ @Arun I have updated my answer. Hope it helps. $\endgroup$
    – Bob D
    Jan 23 '20 at 12:52

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