FOCUSED REVIEW
Focused Review — Thermodynamic Processes and PV Diagrams — Calculus-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Thermodynamic Processes and PV Diagrams.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
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-U09 | TOPIC: Thermodynamic Processes and PV Diagrams | 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 15 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
Key Ideas
Write the differential constraints.
State the constraint for isochoric, isobaric, isothermal, and adiabatic processes using differentials or heat transfer.
Reveal Answers
Isochoric: dV = 0. Isobaric: dP = 0. Isothermal: dT = 0. Adiabatic: δQ = 0.
Why it works: These are local statements of the process constraints.
ACTIVITY 2
Common Mistakes
Recall the P–V work integral and its sign.
Write W for a quasistatic path from V₁ to V₂. What sign results for expansion?
Reveal Answers
W = ∫ from V₁ to V₂ P(V) dV. With positive pressure, expansion V₂ > V₁ gives W > 0.
Why it works: The integration direction follows the change in volume.
ACTIVITY 3
Quick Application
Connect cycle geometry to a closed integral.
What integral represents net work over a complete P–V cycle?
Reveal Answers
Wcycle = ∮ P dV.
Why it works: A closed path accumulates the signed area enclosed by the loop.
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
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Process constraints in differential form
A process constraint restricts the allowed changes of the state variables. For a simple compressible system, the differential first law can be written dU = δQ − P dV when W is work done by the gas. Isochoric, isobaric, isothermal, and adiabatic constraints then simplify different terms or relations.
dU = δQ − P dV; dV = 0, dP = 0, dT = 0, or δQ = 0 by process
Example: For an isochoric process, dV = 0, so the P dV work term vanishes.
Sensei note: Heat δQ and work P dV are path-dependent transfers, not state-function differentials like dU.
KEY CONCEPT 2
P–V work as a path integral
For a quasistatic path, work by the gas is obtained by integrating pressure over volume. The integral is geometrically the signed area under the curve. For a complete cycle, the closed integral gives the signed enclosed area.
W = ∫ P(V)dV; Wcycle = ∮ P dV
Example: If P changes linearly from 3.0 × 10⁵ Pa to 2.0 × 10⁵ Pa while V changes from 2.0 L to 4.0 L, W = 500 J.
Sensei note: Endpoint data alone are not enough unless the process equation P(V) is known.
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
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Guided Example
Integrate a linear P–V path.
Pressure decreases linearly from 3.0 × 10⁵ Pa at 2.0 L to 2.0 × 10⁵ Pa at 4.0 L. Evaluate W.
Reveal Answers
For a linear path, W = PavgΔV = [(3.0 + 2.0)/2 × 10⁵](2.0 × 10⁻³) = 500 J.
Why it works: The integral is the trapezoid area under the straight-line P(V) path.
PRACTICE 2
Independent Check
Use the ideal-gas isothermal path.
One mole of ideal gas expands isothermally at 300 K from 10 L to 20 L. Find W using R = 8.314 J/(mol·K).
Reveal Answers
W = nRT ln(V₂/V₁) = (1)(8.314)(300)ln 2 ≈ 1.73 kJ.
Why it works: For an isothermal ideal gas, P = nRT/V, so the P–V integral produces a logarithm.
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
Compression sign
Use the integration limits.
For an isothermal ideal-gas compression from V₁ to V₂ = V₁/2, is W by the gas positive or negative?
Reveal Answers
Negative: W = nRT ln(1/2) < 0.
Why it works: Compression has V₂ < V₁, making the logarithm and the work by the gas negative.
QUICK CHECK 2
Closed-path work
Interpret the closed integral.
A P–V loop is clockwise. What is the sign of ∮ P dV?
Reveal Answers
Positive.
Why it works: The clockwise orientation gives positive signed enclosed area for work by the gas.
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
Constraints define the path
Use dV = 0, dP = 0, dT = 0, or δQ = 0 to identify how the process is restricted.
KEY TAKEAWAY 2
Integrate pressure over volume
For quasistatic work, W = ∫P dV; for a cycle, ∮P dV gives the signed enclosed area.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
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