How is the work done by a system turned into internal energy according to the first law of thermodynamics?

Question

I have several questions about the first law of thermodynamics:

When we have a force over a piston of a recipient with an ideal gas inside and that piston moves, we have work associated with that. What I can't figure out exactly is how the energy is transferred between the external enviroment and the recipient. I mean, when we have a expansion, the entity that spends energy is our ideal gas, however, we use the external force to calculate work (using the conventional form $\mathrm{d}U = \mathrm{d}W + \mathrm{d}Q$).

So, I'm interested in the mechanism, as some guy explained here: Why does compressing a piston increase the internal energy? (which doesn't completely clearify my doubts)

Answer 1

First law of thermodynamics is the extension of Law of Conservation of Energy for non-isolated system.

There are two forms of first law of thermodynamics(both are actually same):

1. Followed by physicists best suited for dealing with heat-engines: $$\partial E = \partial q - \partial w$$. Here $\partial w$ is the work done by the system.

2. Followed by chemists $$\partial E = \partial q + \partial w$$. Here $\partial w$ is the work done on the system.

---

When the external environment does work, it is actually decreasing the volume of the ideal gas; the molecules collide faster and faster and that is implicit that the kinetic energy is increasing. This topic is greatly explained by Feynman in his lecture.

There is a difference between heat & work. While both refer to energy in transit, the former is defined as the spontaneous flow of energy from one object to other due to temperature difference, while the later is non-spontaneous one; there is always some agent putting energy into the system and hence work doesn't happen automatically.

---

Reply to the comment:

How is energy gained by the molecules ?

The surrounding is doing work on the piston. The piston implies the same to the molecules as a result of which the temperature increases. Now assuming friction & turbulence are negligible, this work will increase the temperature of the gas molecules. Now, the way they gain energy is not only due to change in kinetic energy(that you associated with change in linear momentum), but also rotational energy & vibrational energy. Every molecule has certain degrees of freedom in their motion: the number of independent ways there can be motion: translational , rotational , vibrational. For instance, at a certain temperature, a monoatomic molecule can move along three coordinates. A diatomic molecule can even rotate along two axes. The energy associated can be given by: $$\dfrac{1}{2} m{v_x}^2 \quad, \dfrac{1}{2} m{v_y}^2 \quad, \dfrac{1}{2} m{v_z}^2 \quad, \dfrac{1}{2} I {\omega_x}^2 \quad, \dfrac{1}{2} I {\omega_y}^2$$. Each form of energy is a quadratic function of a coordinate or velocity component. Each degree of freedom act as a storage of energy. So, when the piston does work, the gas gain the energy by actually activating each degrees of freedom. And at temperature $T$, the average energy of any quadratic degree of freedom is $\dfrac{1}{2} kT$. This is known as Equipartition Theorem.

Answer 2

Consider a container containing n moles of an ideal gas. The gas exerts a pressure P on the container and the piston. If P equals the atmospheric pressure, then the piston does not move, as it experiences equal forces from in and out of the container.

When you increase the external pressure, the gas in the container is compressed. If the compression of the gas is isothermal (the temperature of the gas doesn't change during the process) then internal energy doesn't change.

Consider the ideal gas equation. $$PV = nRT$$ If the compression of the gas is isobaric (no change in pressure), then the temperature of the gas will definitely increase, causing it's internal energy to increase.

An adiabatic process, where $\Delta Q = 0$, is a bit more tricky to deal with. Both the pressure and temperature change on compression. Using the fact that $PV^{\gamma}$ and $\frac{PV}{T}$ are constants, we divide to say that $V^{\gamma-1}T$ is also a constant. So clearly decreasing the volume will increase the temperature. ($\gamma$ is the ratio of the molar specific heat capacities for a fixed pressure to a fixed volume $\gamma =\frac{ C_p}{C_v}$ It is always grater than one.)

Note that I keep taking increase in temperature as increase in kinetic energy. There is a formula that relates T to internal kinetic energy of the gas particles. For a monoatomic gas, the kinetic energy is $\frac{3}{2}KT$. $K$ is the Boltzmann constant.