1.7.13 · D5Thermodynamics
Question bank — First law of thermodynamics — dU = dQ − dW, sign conventions
Figures — what path dependence and sign zones look like
Work is the area under the – curve; two paths between the same endpoints enclose different areas — that is why and are path functions.
The sign-convention zones: expansion (area swept to the right, ) versus compression (swept left, ); heat in versus heat out .
Why is path-dependent: two routes A→B enclose an area = the net work, and by that different forces a different even though is identical.
True or false — justify
Heat is a form of energy stored inside a gas.
False. Heat is energy in transit driven by a temperature difference; once it enters, it becomes internal energy — you cannot point at "the heat inside" a gas.
If two processes start and end at the same , they must transfer the same .
False. is a path function. Only (state function) is fixed by the endpoints; and can differ wildly while stays equal — see the two-path figure above.
If two processes have the same endpoints, they have the same .
True. is a state function, so depends only on the endpoints, never the path.
In physics convention, compressing a gas gives positive .
False. is work done by the gas; compression means , so . The energy you pump in appears as in .
For an ideal gas, adding heat always raises its temperature.
False. In isothermal expansion the gas absorbs heat yet is constant — all the heat leaves again as work ().
The First Law tells you which direction a process will go.
False. The First Law only enforces energy bookkeeping; direction (irreversibility) is the job of the Second law of thermodynamics.
only holds for ideal gases.
False. It is pure energy conservation and holds for any closed system — solids, liquids, real gases too. Only the shortcut needs the ideal-gas assumption.
In a free expansion into vacuum, the gas does positive work because it expands.
False. Work ; the external pressure is zero (vacuum), so despite the volume increasing.
The only kind of work in the First Law is pressure–volume ("") work.
False. is all work the system does: electrical work on a battery, magnetic work, surface-tension work on a stretched film, shaft work, etc. For a simple gas the only channel is , but the law covers every mode.
Chemistry's contradicts physics' .
False. Same physics, different : chemistry's is work done on the gas , so both give identical .
An adiabatic process must be fast.
Mostly true, but the real requirement is . Fast processes leave no time for heat flow, and perfect insulation also gives — either route makes it adiabatic.
Spot the error
"The gas absorbed 500 J of heat, so its internal energy rose by 500 J."
Error: ignores work. ; if the gas also did 200 J of work, only 300 J stays as .
"Isothermal means no heat flows, since temperature doesn't change."
Error: confuses isothermal with adiabatic. Isothermal keeps fixed but heat does flow (in = out via work). Adiabatic is the one with .
"Since is a state function, and must be too — they're all energies."
Error: being an energy doesn't make something a state function. describe transfers along a path; only the stored quantity is a state function.
"Compression pumps energy in, so must be positive."
Error: not necessarily. If the gas also loses more heat than the work put in (, ), can be negative. You must add both channels: .
"For an ideal gas at constant volume, because no work is done."
Error: constant volume means , not . Here , so all heat raises internal energy — heat definitely flows.
" works for every process."
Error: only when is constant (isobaric). In general , and the pressure varies along isothermal/adiabatic paths.
"In an adiabatic expansion the gas cools, so it must have released heat."
Error: adiabatic means — no heat released at all. It cools because it spends internal energy doing work: .
"Work by the gas is negative during expansion because the gas loses energy."
Error: losing energy is real, but the sign convention defines expansion () as . The energy loss shows up as the term reducing , not as a negative .
Why questions
Why is a tiny bit of heat written while a tiny bit of internal energy is written ?
Because has a well-defined value at each state, so integrates to ; there is no "heat content" of a state, so (with the inexact flag) depends on the path and cannot be integrated from endpoints alone. Look at the two-path figure: same endpoints, different .
Why does an ideal gas's internal energy depend only on temperature, not pressure or volume?
Ideal-gas molecules have no intermolecular potential energy, so is purely kinetic; average kinetic energy is set only by (see Internal energy and degrees of freedom).
Why is for an isothermal ideal-gas process?
Because only, and is held constant, so cannot change even though heat and work both flow.
Why must we ask "work by or on the gas?" before plugging signs?
Because the same physical situation gives opposite-signed depending on the convention; the physics formula needs work by the gas. The sign-zone figure shows expansion as "" and compression as "".
Why does and not ?
Work is force times distance; the gas force on the piston is and the distance swept gives , so — geometrically, the thin sliver of area under the – curve. A change in with fixed moves no piston, hence no area, hence no work.
Why can heat added to a gas sometimes leave the internal energy unchanged?
If the gas simultaneously does an equal amount of work (), then — exactly the isothermal case.
Why must we integrate the external pressure , not the gas pressure, for work?
Work is done against whatever the piston actually pushes on — the surroundings. Only in a slow (reversible) process is the gas pressure equal to the external pressure at every instant; in a sudden expansion the gas pressure can exceed , and the real work is the smaller .
Why is the First Law useless for predicting whether ice melts spontaneously?
It only checks that energy balances; it never forbids energy-conserving processes that run "backwards". Spontaneity is governed by entropy in the Second law of thermodynamics.
Edge cases
Free expansion into vacuum for an ideal gas: what are , , , ?
(external pressure is zero), (fast, insulated), so and — temperature is unchanged even though volume jumps.
A cyclic process returns to its start: what is over one full cycle?
Zero, because is a state function and the start and end states are identical; therefore over the cycle — the area enclosed by the loop on the – diagram.
Isochoric (constant-volume) heating: which term vanishes and what remains?
since , so — every joule of heat becomes internal energy.
Adiabatic process with : does that mean ?
No. gives , so the internal energy changes by exactly minus the work done by the gas — it heats on compression, cools on expansion.
A gas expands at constant temperature but is a real gas with attractions: is still zero?
Not necessarily. Real gases have intermolecular potential energy that changes with volume, so depends on too; can be nonzero even at constant .
A battery inside the system delivers electrical energy: does the First Law still hold?
Yes — but you must count electrical work as a work mode: . The law never assumed work was only .
Zero heat and zero work (, ): what can you conclude?
— the internal energy (and, for an ideal gas, the temperature) is unchanged; the state's energy is completely isolated.
Constant-pressure heating of an ideal gas: why does exceed ?
Because the gas also expands and does work , so ; the extra heat pays for the work (this is why , see Heat capacities Cp and Cv).
Compressing a gas while it sheds exactly as much heat as the work put in: what is ?
Zero. Work on the gas raises energy, heat loss removes an equal amount: .
Connections
- Work done in thermodynamic processes (PV diagrams)
- Isothermal, adiabatic, isobaric, isochoric processes
- Internal energy and degrees of freedom
- Heat capacities Cp and Cv
- Conservation of energy (mechanics)
- Second law of thermodynamics