FULL REVIEW
Full Review — First Law of Thermodynamics — Calculus-Based
Review the essential ideas, relationships, and problem-solving tools for First Law of Thermodynamics.
TIME
45–60 minutes
BEST FOR
A complete unit review
FINISH WITH
A readiness check
After this full review, you'll be able to...
recall the essential ideas, apply them to representative problems, and determine what to study next.
Choose how you want to review
Course Alignment
This Physics Sensei Unit Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
RESOURCE: Physics Sensei Unit Review | UNIT ID: THM-U08 | TOPIC: First Law of Thermodynamics | COURSE LEVEL: Calculus-Based
BEST USED ✓ After learning the unit ✓ Before starting homework ✓ Before a quiz or exam
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 45–60 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Recall Activity 1
Differential form
Write the first law for an infinitesimal change using the work-by-system convention.
Reveal Answers
dU = δQ - δW.
Why it works: U is a state function, while heat and work are path-dependent transfers.
ACTIVITY 2
Recall Activity 2
Boundary work
Write the quasistatic pressure-volume work element.
Reveal Answers
δW = P dV.
Why it works: Positive dV gives positive work by the system.
ACTIVITY 3
Recall Activity 3
Path integral
Write the work for a quasistatic process from initial volume to final volume.
Reveal Answers
W = ∫ P dV from initial volume to final volume.
Why it works: The value depends on the path P(V), not only the endpoints.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
KEY CONCEPT 1
First Law and Exact Differentials
For the work-by-system convention, dU = δQ - δW and over a finite process ΔU = Q - W. Because U is a state function, dU is exact; heat and work are process-dependent and are commonly written with inexact differentials.
dU = δQ − δW and ΔU = Q − W
Example: For a complete cycle, ΔU = 0, so net Q equals net W.
Sensei note: Keep the sign convention fixed throughout a derivation.
KEY CONCEPT 2
Path Dependence and State Functions
Between fixed equilibrium states, ΔU is unique. Q and W may change with the path while their difference remains ΔU. This is central to interpreting P-V diagrams and thermodynamic cycles.
ΔU = ∫ dU = U₂ − U₁
Example: Different paths between the same endpoints can enclose different areas and therefore produce different work.
Sensei note: Endpoint data alone determine ΔU but generally not Q or W separately.
KEY CONCEPT 3
Quasistatic PV Work
For simple compressible work, δW = external pressure dV; in a quasistatic process external pressure is effectively the boundary pressure and W = ∫ P dV. Constant pressure reduces to W = P(final volume - initial volume).
δW = P dV and W = ∫ P dV
Example: If P(V) = a/V, then W = a ln(final volume/initial volume).
Sensei note: Use consistent SI units so pressure-volume integrals return joules.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
PRACTICE 1
Worked Example
A system absorbs 1.60 kJ and performs 0.55 kJ of work. Find ΔU.
Apply the finite first-law balance.
Reveal Answers
ΔU = 1.60 kJ - 0.55 kJ = 1.05 kJ.
Why it works: This is the integral form of dU = δQ - δW.
PRACTICE 2
Guided Problem
A quasistatic expansion follows P(V) = a/V with a = 500 J from initial volume = 2.0 L to final volume = 5.0 L. Find W.
Integrate W = ∫(a/V)dV = a ln(final volume/initial volume).
Reveal Answers
W = 500 ln(5/2) ≈ 458 J.
Why it works: The logarithm uses a dimensionless volume ratio.
PRACTICE 3
Independent Problem
For the expansion above, ΔU = 120 J. Find Q.
Use Q = ΔU + W after evaluating the path work.
Reveal Answers
Q ≈ 120 J + 458 J = 578 J.
Why it works: Path work plus the state-function change determines the required heat transfer.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Cycle relation
Solve using the first law and the stated sign convention.
For a complete cycle, what relation connects net heat and net work?
Reveal Answers
net Q = net W.
Why it works: Because ΔU for the cycle = 0.
QUICK CHECK 2
Integral work
Solve using the first law and the stated sign convention.
For constant P = 3.0 × 10⁵ Pa and ΔV = 1.5 × 10⁻³ m³, evaluate ∫P dV.
Reveal Answers
W = 450 J.
Why it works: Constant P factors out of the integral.
QUICK CHECK 3
Solve for heat
Solve using the first law and the stated sign convention.
A path has W = 620 J and ΔU = -140 J. Find Q.
Reveal Answers
Q = 480 J.
Why it works: Q = ΔU + W = -140 + 620.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Differential first law
dU = δQ - δW; over a finite process, ΔU = Q - W.
KEY TAKEAWAY 2
Path dependence
U is a state function; heat and work depend on the path.
KEY TAKEAWAY 3
PV integral
For quasistatic work, W = ∫ P dV, with constant pressure as a special case.
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Next Step
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