MCAT content library / Chemistry & physics / The first law
Choose a system and sign convention before combining heat and work, then use a pressure–volume path to separate state changes from path-dependent energy transfer.
About 17 minutes, including a two-path PV example and checks.
Useful foundations
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Define the system before writing the energy balance
The first law is energy conservation applied to a thermodynamic system. This lesson uses the chemistry convention:
ΔU = q + w
where q is heat transferred into the system and w is work done on the system. Thus q > 0 when the system absorbs heat. Expansion against an external pressure does work on the surroundings, so pressure–volume work on the system is negative:
w = −∫PextdV.
A physics text may instead write the first law using work done by the system. Either convention is valid if used consistently.
Concept: internal energy is a state function
Internal energy U depends on the thermodynamic state. If two processes begin at the same initial state and end at the same final state, they have the same ΔU even if they exchange different amounts of heat and work along the way.
Heat q and work w are not state functions. They describe energy crossing the system boundary through different mechanisms. Different paths between the same endpoints can therefore have different q and w while preserving the same sum q + w.
Worked application: two paths, same endpoints
An original gas process moves from state I to state F, expanding from 1.0 L to 3.0 L. The two paths share the same initial and final states, so the measured internal-energy change is the same for both: ΔU = +100 J.
| Path | Expansion pressure | Work by gas | w on system | Required q |
|---|---|---|---|---|
| A | 200 kPa | +400 J | −400 J | +500 J |
| B | 100 kPa | +200 J | −200 J | +300 J |
For Path A, w = −(200 kPa)(2.0 L) = −400 J. The first law gives 100 J = q − 400 J, so q = +500 J.
For Path B, w = −200 J, so 100 J = q − 200 J and q = +300 J. Both processes reach the same ΔU, but the higher-pressure expansion transfers more energy as work and therefore requires more heat input to reach the same final state.
Pattern: PV work is area under the path
On a pressure-versus-volume graph, the magnitude of quasistatic or specified external-pressure work is related to the area under the process path. For a constant pressure, the area is a rectangle PΔV. For a changing pressure, the integral adds the narrow pressure-by-volume contributions along the path.
A path that lies at higher pressure during the same expansion generally has a larger work-by-system area. The sign still depends on which convention is being used.
Rule: heat and work depend on the path
If a problem tells you only the initial and final states, you may be able to determine ΔU from state information, but you cannot generally determine q and w separately without knowing the process path or another constraint.
This is why “same endpoints” does not imply “same heat.” It implies the same change in state functions such as internal energy. Heat and work adjust to satisfy energy conservation for the particular route taken.
ACTIVE PRACTICE · THERMODYNAMIC ACCOUNTING
Keep the Endpoints Fixed While Work Changes
The original gas expands from 1.0 to 3.0 L with the same initial and final states along two stipulated constant-pressure paths. Both paths have ΔU = +100 J; by the chemistry convention ΔU = q + won, and won = −PextΔV for these paths. One kPa·L equals one joule.
Volume change = +2.0 L. Work done by the system = J. Work done on the system = J.
Heat input required to produce ΔU = +100 J: J.
The demonstration uses the lesson’s stipulated external-pressure work relation. Pressure may differ between paths while the initial and final states are held fixed; it is not claiming all paths are equilibrium constant-pressure processes.
Check your reasoning
Choose an answer before opening its feedback.
Check 1
Using ΔU = q + w with w defined as work on the system, a gas expands and does 250 J of work on the surroundings while absorbing 400 J of heat. What is ΔU?
A. +150 J
Correct. w = −250 J, so ΔU = 400 − 250 = 150 J.
B. +650 J
Review this choice. Expansion work on the system is negative in the stated convention.
C. −150 J
Review this choice. Heat input exceeds the magnitude of expansion work.
D. −650 J
Review this choice. Both signs are mishandled.
Check 2
Two paths connect the same initial and final equilibrium states. Which quantity must be identical for both paths?
A. ΔU
Correct. Internal energy is a state function.
B. q
Review this choice. Heat is path dependent.
C. w
Review this choice. Work is path dependent.
D. The area under every PV path
Review this choice. Different paths can enclose different work areas.
Check 3
A gas expands at constant external pressure from 2 L to 5 L. If the pressure is doubled while the volume change is unchanged, what happens to the magnitude of PV work?
A. It doubles.
Correct. |w| = P|ΔV| for constant pressure.
B. It halves.
Review this choice. Work magnitude is directly proportional to pressure.
C. It stays the same.
Review this choice. Pressure has changed.
D. It becomes zero.
Review this choice. There is still a nonzero pressure and volume change.