# In an adiabatic process, why is it that if work is done very quickly then there is no time for energy transfer to the surrounding?

I don't seem to understand the logic in adiabatic process of how 'if work is done on the system very quickly e.g. compress a piston containing n number of gas particles, there is no time for energy to transfer to the surrounding'. However, temperature of the gas in the system will change in the process, this implies that if work is done very quickly, then all the energy from work done will have time to transfer to the kinetic energy gained by the gas; but there won't be enough time for heat to transfer to the surrounding.

In the example I proposed, heat transfer to the gas particles to increase their KE will occur by conduction through energy conducting throughout the gas particles (please correct me if I'm wrong). Thus if work is done very quickly, why is there time for work done on gas to transfer heat to the gas particles, but no time for heat to transfer to the surrounding by transferring heat through the piston walls (assuming the walls are not insulated for this adiabatic process)?

• For a 4 stroke car engine running at 2500 rpm, the power stroke occurs in 1/2 of a cycle. There IS heat transfer during this part of the cycle, but the amount of PV work done by the piston is WAY WAY larger than the amount of heat lost during the time that this half-cycle is happening. From an engineering standpoint, the amount of heat transfer is insignificant when compared to the amount of work done, so it is ignored. – David White Oct 21 '18 at 1:06

There is a rate at which an object cools which depends on the temperature difference between system and surroundings. This is Newton's law of cooling.

$$\frac{dQ}{dt} \propto (T_{system}-T_{surroundings})$$

When a process occurs so rapidly that the work done on the system is much larger than the heat transfer between the system and surroundings in that time interval, the process can be considered to be adiabatic.

Using first law of thermodynamics

$$dQ =dU -dW$$

Negative work is done because work is done on the system

$$0 =dU - dW$$

$$dU = dW$$

Since internal energy of ideal gas depends only on temperature. Therefore, temperature of working fluid i.e. gas increases.

As I understand your question, you're asking "Why would energy added to a system by either work or heat arrive at different rates? In other words, why can the latter mechanism be suppressed relative to the former mechanism?" The answer is that heat and work operate by fundamentally different mechanisms that are decoupled and can be controlled largely independently. It's useful to look at exactly how these processes operate in a gas cylinder.

To do work on a gas in the compression manner you describe, you would press on a piston handle, transmitting energy to the inner surface of the piston through a pressure wave moving at the speed of sound in the piston material. Since the bulk modulus of solids is far greater than the bulk modulus of the typical gas, it's reasonable to assume that the piston doesn't compress at all. The gas in turn interacts with the relatively rigid piston surface, with gas molecules that contact the moving piston getting a momentum boost in the opposite direction. The price of this momentum boost is the resistance you'd feel at the handle.

In contrast, heating the gas requires diffusive energy transport through the walls and piston, which is a completely different mechanism. This process relies on individual molecules randomly pushing against each other and has no well-defined speed (in fact, the effective speed scales inversely with the wall thickness, which suppresses heat transfer over large distances even though mechanical work is largely unaffected). One consequence is that it's straightforward to incorporate a mechanically stiff structure with a low thermal conductivity. One example would be to use metal struts/meshwork separated by insulation, air, or even vacuum, as is done in thermos bottles.

You ask "why is there time for work done on gas to transfer heat to the gas particles"

The short and simple way to summarize Mechanic7's excellent answer is this: Work done on the gas particles is what creates the heat. When a gas is compressed, it heats up. If there is no (or very little) heat exchange with the surroundings, the process is considered adiabatic. The heat does not have to be transferred to the gas particles, it is created there by the compression.

• A body has no "heat", a body has "energy" and "entropy". "Heat" is not created, "heat" is not a thing; "heat" is energy and entropy being transferred, ie., "heating" from one place to another. – hyportnex Oct 21 '18 at 14:04
• @hyportnex Thanks for your input. So just to confirm, when work is done on the gas particles, energy is indeed transferred from the piston's movable lid to the gas particles inside the piston? And I believe this mechanism of heat transfer is conduction? – Bøbby Leung Oct 21 '18 at 17:21
• When work is done in an adiabatic process by pressing on a piston there is work done and internal energy increases but no heat is transferred, if the process is reversible then while the internal will have increased the entropy stays the same. If you allow for thermal conduction then both energy and entropy will change even if it is internally (aside from the conduction through walls) a reversible process. – hyportnex Oct 21 '18 at 17:28