1.7.14 · D5Thermodynamics
Question bank — Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)
Before diving into the traps, let's earn every symbol we'll lean on — so no reveal below ever surprises you.
The four processes are just four paths between states. This picture is the map for the whole page:
Recall The two facts that resolve almost every trap
- in every process (internal energy of an ideal gas depends only on temperature — see the First Law of Thermodynamics).
- The First Law is always true; the four named processes just each kill one term or one variable.
True or false — justify
Each statement is followed by the honest reasoning, not a bare verdict.
Isothermal means no heat flows.
False. Isothermal fixes temperature, so and the First Law forces — heat flows freely to hold constant. The zero-heat process is adiabatic.
In an adiabatic process the temperature stays constant.
False. Adiabatic fixes , not . With no heat in, any work done by the gas is paid from internal energy, so drops on expansion and rises on compression.
Because is the "constant-volume" heat capacity, only holds at constant volume.
False. depends only on , so this relation holds on every path. is merely the conversion factor between and ; the "V" in its name comes from how it is measured, not where it applies.
In an isochoric process the gas does no work, so no energy enters it.
False. Work is zero, but heat is not: . Every joule of heat goes into internal energy since none is spent on expansion.
The isotherm and the adiabat through the same point have the same shape.
False. Isotherm is ; adiabat is with . The adiabat is steeper, dropping to a lower pressure for the same volume increase.
For an isobaric expansion, because pressure is constant.
False. At constant the gas also does work , so , strictly larger than .
Mayer's relation only applies to the isobaric process where it was derived.
False. It relates two material properties of the gas; once proven it holds always. We derived it using an isobaric process, but the conclusion is path-independent.
If a gas returns to its starting , , and , then for the round trip.
False. over a closed loop, so — and equals the enclosed loop area, generally nonzero. Only is guaranteed zero.
A vertical line on a – diagram represents an isobaric process.
False. Vertical means fixed → isochoric. Horizontal (constant height ) is isobaric.
Work done by the gas is always positive when the gas gains heat.
False. In isochoric heating but ; in adiabatic compression with . Sign of tracks volume change, not heat.
Spot the error
Read the flawed line, then say what breaks.
"Adiabatic: ; the gas was compressed so , and I got ."
The sign is wrong. With , the factor , so — work is done on the gas, which is exactly why compression heats it.
"Isothermal work: ."
The ratio is flipped. Correct is ; expansion () must give , but the flipped version would give a negative number.
"Since in an isothermal process, and , the gas has no heat capacity here."
The reasoning conflates with " vanishes." only because ; is a fixed property of the gas and is unchanged.
"For adiabatic I'll use to find the end state."
Wrong curve. Adiabatic obeys ; is the isotherm. Using it underestimates how far the pressure falls.
"In isobaric expansion (pressure times volume change)."
Variables are swapped. Pressure is the constant one; (constant pressure times volume change), the area of a rectangle of height .
" works for any heating, since is just the heat capacity."
is the isobaric result. Heat is path-dependent: for isochoric it's , for isothermal it's . You cannot pick one for all paths.
"Adiabat and isotherm both satisfy ."
Only the isotherm does (). The adiabat has instead, and in general because temperature changes.
Why questions
The "why" is where memorised formulas either survive or collapse.
Why is larger than ?
At constant pressure the gas expands as it warms, so incoming heat must pay for both the temperature rise and the expansion work; Mayer's relation quantifies the gap as exactly .
Why does an adiabatically expanding gas cool even though nothing "takes" heat away?
With , the work of pushing outward can only be financed by the gas's own internal energy; lowering lowers . This is why rising air forms clouds and why a spray can gets cold.
Why is work equal to the area under the – curve?
Each tiny expansion against pressure contributes ; summing (integrating) these thin strips is precisely the area beneath the curve (see the shaded strip in the map figure above).
Why does the shape of the – path matter for work but not for ?
is the area under the path, so different paths between the same endpoints sweep different areas. depends only on the endpoints' temperatures, so it is path-independent.
Why must an isothermal change happen slowly (quasi-statically)?
"Quasi-statically" means so gently that the gas is in balance at every instant. Only a slow process lets heat trickle in fast enough to keep temperature pinned to the reservoir the whole way; rush it, and parts of the gas would heat or cool before equilibrium restores a single .
Why must an adiabatic change be either well-insulated or very fast?
"Well-insulated" means wrapped so heat physically cannot cross the walls. needs no heat to leak: insulation blocks the path, or extreme speed finishes the process before any meaningful heat has time to cross the boundary.
Why is the adiabat steeper than the isotherm at the same point?
On the adiabat the gas also loses temperature as it expands, so pressure falls for two compounding reasons (more volume and less ); the exponent encodes this extra steepness.
Why does let us find internal energy change without knowing the path?
Because is a state function of alone for an ideal gas — pick any imaginary path with the same and you get the same .
Edge cases
Boundary and degenerate scenarios the topic quietly assumes.
What is the work in an isothermal process where the volume does not change at all?
, and then too — the "process" is just the gas sitting still.
For an isobaric process at constant temperature, what happens?
If both and are fixed, then forces fixed as well — nothing changes; it is a single equilibrium point, not a process.
Can a process be simultaneously isothermal and adiabatic (for a real change of state)?
Only trivially. Isothermal needs ; adiabatic needs ; together they force and no volume change, i.e. no actual process.
In the limit , what does the adiabat become?
, so the adiabat collapses onto the isotherm. Physically means expansion barely changes , blurring the two curves.
If a gas is compressed adiabatically to zero net temperature change, is that possible?
No. Adiabatic compression does positive work on the gas with , so and must rise; zero would require , contradicting .
What is for a complete cyclic process (a closed loop on the – diagram)?
Exactly zero, because the gas returns to its initial temperature. The net heat then equals the net work, which is the enclosed area — the basis of engines like the Carnot Cycle.
For an isochoric process, is any of the added heat "wasted" as work?
None — with the gas does zero work, so of the heat becomes internal energy. It is the most "efficient" at raising temperature per joule.