# Thermodynamical Work: Understanding Pext, Pint & dV

• emailanmol
In summary, the first law of thermodynamics states that the heat added to a system is equal to the change in internal energy plus the work done on the system. Work can be defined as either the work done by the surrounding on the gas or the work done by the gas on the surrounding, but these two are only equivalent for reversible processes. Additionally, in order to calculate work, it is important to use the correct convention for the pressure, either external or internal. Finally, according to Newton's third law, action and reaction are equal and opposite, so when calculating work, only one of the two should be used.
emailanmol
I seem to have a conceptual doubt.

In thermodynamics first law
Q + W= U
W is the work done by the surrounding on gas and has value integral of [-P(external)dV].
Where dV is change in volume of gas

Now in some other conventions , we define the 1st law as Q-W = U
Where W is work done by gas on surrounding.

My question is.

How Is work done by surrounding on gas = -(Work done by gas on surrounding)

According to me work by gas on surrounding is integral of [P(internal)dV]

So these two are equivalent only for reversible processes where Pext is nearly equal to Pint.

Also whenever a gas compresses due to external pressure, the surrounding does work on gas and gas gains energy.But while getting compressed the gas does work done surrounding too and loses some of its energy.(Am i right?)
So isn't the net work done on gas a difference between the two i.e isn't it w=-(Pext -Pint)dV

Where am I going wrong?
It will be really kind of you to be descriptive and to help me identifying the mistakes in my analogy, and the required corrections.
Thanks

emailanmol said:
I seem to have a conceptual doubt.

In thermodynamics first law
Q + W= U
W is the work done by the surrounding on gas and has value integral of [-P(external)dV].
Where dV is change in volume of gas

Now in some other conventions , we define the 1st law as Q-W = U
Where W is work done by gas on surrounding.

My question is.

How Is work done by surrounding on gas = -(Work done by gas on surrounding)

According to me work by gas on surrounding is integral of [P(internal)dV]

So these two are equivalent only for reversible processes where Pext is nearly equal to Pint.

That's entirely true. You can only replace P(ext) by -P(int) if they are in equilibrium .
To the second part of your question: According to Newton actio=reactio always and you have to use one of the two (usually actio) to calculate work.

Hey thanks.Now its clear

## 1. What is thermodynamical work?

Thermodynamical work is the transfer of energy between a system and its surroundings through a change in volume or pressure. It is a fundamental concept in thermodynamics and plays a crucial role in understanding the behavior of physical systems.

## 2. How is Pext different from Pint in thermodynamics?

Pext, or external pressure, refers to the pressure applied by the surroundings on the system, while Pint, or internal pressure, refers to the pressure within the system itself. In thermodynamics, the difference between Pext and Pint is important in determining the direction and magnitude of thermodynamic work.

## 3. What is the relationship between Pext, Pint, and dV in thermodynamics?

In thermodynamics, Pext and Pint are related through the equation Pext = Pint + dP, where dP represents the change in pressure. This relationship is important in understanding how work is performed on a system by the surroundings, and how the system responds to this work.

## 4. How does thermodynamical work relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. Thermodynamical work is a form of energy transfer, and is therefore closely related to the first law of thermodynamics. The work done on a system can be seen as a change in the system's internal energy, as described by the first law.

## 5. Can thermodynamical work be negative?

Yes, thermodynamical work can be negative. This occurs when the system does work on the surroundings, rather than the other way around. In this case, the system's internal energy decreases and the surroundings' increases, resulting in a negative value for thermodynamical work.

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