# Comparison between isobaric, isothermal and adiabatic expansion

Please check if my understandings of these terms are right because they are a bit confusing to me.

For all three types of expansions, a change in volume means work is done by the system to the surrounding.

Isobaric expansion: As $p = F/A$ applies to the force of piston over area. The system is in isobaric equilibrium as long as the force does not change. In order for volume to increase, heat must enters the system to increase internal energy, leading to increase in volume.

Isothermal expansion: takes in heat from a heat reservoir, convert that heat into work energy and the system expands in volume. As temperature must remain constant, the heat energy absorbed is converted into work energy instead of internal energy. (But if heat is converted to internal energy, wouldn't the system increase in volume as well?)

Adiabatic expansion: As $Q= 0$, work should be constant. In this case, isn't volume will remain constant as well so no expansion will take place?

• "But if heat is converted to internal energy, wouldn't the system increase in volume as well?" If there is no space to expand at, the pressure just increases. – Steeven Nov 9 '14 at 21:52
• Isothermal simply means a process at constant temperature. Think about ice melting at $0^oC$. A lot of heat is added to melt it, but the temperature is 0. – Steeven Nov 9 '14 at 21:54

Yes quite! The ideal gas law (and in general any state equation) holds only on equillibrium while the 1st law (and all the rest) hold in general. So, your anslysis mr James Hoyland is inacurrate.

Mr Steven, the post includes the word "expansion" so volume changes by assumption.

And mr or ms PhysC, the first two cases are correct. About the third, expansion would occur by pulling the piston thus removing energy from the system and causing decrease on temperature

Not quite! Using the ideal gas law: $\frac{pV}{T}=constant$ and the First Law of Thermodynamics $dU = \delta W + \delta Q$:

For an isobaric change pressure remains constant so if volume increases then temperature must increase as well. This implies the internal energy must increase. However there is work done either by or on the system $dW = p\Delta V$. If this was an expansion so volume increases then the system must do work, without a heat transfer this would make the the internal energy go down in accordance with the first law so there must also be a heat flow to make sure the net internal energy goes up.

In isothermal expansion we keep temperature the same so if volume increases then pressure must go down. We achieve this by keeping the piston in contact with a large heat reservoir. On the pV diagram the isothermal change then has the form $p \propto V^{-1}$ so the work is the integral of this. However , in this case there must be an exchange of heat also to keep the temperature constant.

For an adiabatic process there is no heat flow, however if there is a change in volume there must be work involved. For this we also need to use the adiabatic relationship which is that $pV^{\gamma}=constant$ where $\gamma$ is the ratio of specific heats of the gas. Pressure, Temperature and Volume all change in this case.

## protected by AccidentalFourierTransformJul 16 '18 at 20:49

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