FULL REVIEW
Full Review — Thermodynamic Processes and PV Diagrams — Calculus-Based
Review the essential ideas, relationships, and problem-solving tools for Thermodynamic Processes and PV Diagrams.
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-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 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
Recall the differential constraints for the four common processes.
State the local constraint for isochoric, isobaric, isothermal, and adiabatic processes.
Reveal Answers
dV = 0, dP = 0, dT = 0, and δQ = 0, respectively.
Why it works: These constraints determine which terms and state relations simplify.
ACTIVITY 2
Recall Activity 2
Write the quasistatic P–V work integral.
Give W by the gas for a path P(V) from V₁ to V₂ and state the sign for ordinary expansion.
Reveal Answers
W = ∫ from V₁ to V₂ P(V) dV. For positive pressure and V₂ > V₁, W is positive.
Why it works: Work is accumulated along the volume coordinate, so the integration direction sets the sign.
ACTIVITY 3
Recall Activity 3
Recall the cycle integral.
Write the expression for net work around a closed P–V cycle and interpret its geometry.
Reveal Answers
Wcycle = ∮ P dV; it is the signed area enclosed by the loop.
Why it works: The contributions from expansion and compression combine into a closed-path area.
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
Differential constraints organize thermodynamic processes
For a simple compressible system using work done by the gas, the differential first law is dU = δQ − P dV. A process constraint then restricts one variable or transfer: dV = 0 for isochoric, dP = 0 for isobaric, dT = 0 for isothermal, and δQ = 0 for adiabatic.
dU = δQ − P dV; dV = 0, dP = 0, dT = 0, or δQ = 0 by process
Example: Along an isochoric path, dV = 0, so the boundary-work contribution P dV vanishes even though pressure and temperature may change.
Sensei note: δQ is written differently from dU because heat is path dependent while internal energy is a state function.
KEY CONCEPT 2
Work is the integral of pressure along the volume path
For a quasistatic process, W depends on P(V) along the entire path. Geometrically, the integral is the signed area under the P–V curve. A curved process cannot generally be replaced by a single pressure times ΔV unless that pressure represents the integral correctly.
W = ∫ P(V)dV
Example: For a linear path from 3.0 × 10⁵ Pa to 2.0 × 10⁵ Pa while volume rises from 2.0 L to 4.0 L, the trapezoid area gives W = 500 J.
Sensei note: Two paths connecting the same end states can have different work because P(V) differs between them.
KEY CONCEPT 3
Ideal-gas paths and closed cycles
For an isothermal ideal-gas process, P = nRT/V, so integration gives a logarithmic work expression. For a closed cycle, the system returns to its initial state and the net work is the closed P–V integral. Clockwise orientation gives positive net work by the gas.
W = nRT ln(V₂/V₁) for isothermal ideal gas; Wcycle = ∮ P dV
Example: One mole at 300 K expanding isothermally from 10 L to 20 L does about 1.73 kJ of work.
Sensei note: The cycle integral depends on orientation; reversing the same loop reverses the sign of the work.
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
Evaluate the integral for a linear pressure path.
Pressure decreases linearly from 3.0 × 10⁵ Pa at 2.0 L to 2.0 × 10⁵ Pa at 4.0 L. Find W.
Reveal Answers
The integral equals the trapezoid area: W = [(3.0 + 2.0)/2 × 10⁵ Pa](2.0 × 10⁻³ m³) = 500 J.
Why it works: For a linear function, the average value over the interval is the arithmetic mean of the endpoint pressures.
PRACTICE 2
Guided Problem
Integrate the isothermal ideal-gas curve.
One mole of ideal gas at 300 K expands isothermally from 10 L to 20 L. Calculate W.
Reveal Answers
W = nRT ln(V₂/V₁) = (1)(8.314)(300)ln 2 ≈ 1.73 × 10³ J.
Why it works: Substituting P = nRT/V into ∫P dV gives the logarithm.
PRACTICE 3
Independent Problem
Use the closed integral for a rectangular cycle.
A clockwise rectangle spans pressures 1.0 × 10⁵ to 3.0 × 10⁵ Pa and volumes 2.0 to 5.0 L. Evaluate Wcycle.
Reveal Answers
Wcycle = enclosed area = (2.0 × 10⁵ Pa)(3.0 × 10⁻³ m³) = +600 J.
Why it works: The vertical legs have dV = 0, and the two horizontal integrals differ by the rectangle area.
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
Linear path integral
Use the trapezoid or integrate explicitly.
P rises linearly from 2.0 × 10⁵ Pa to 4.0 × 10⁵ Pa while V increases from 1.0 L to 3.0 L. Find W.
Reveal Answers
W = PavgΔV = (3.0 × 10⁵)(2.0 × 10⁻³) = 600 J.
Why it works: A linear P(V) integrates to the trapezoid area.
QUICK CHECK 2
Isothermal compression
Use the logarithmic expression.
For an ideal gas compressed isothermally from V₁ to V₁/2, what is the sign of W by the gas?
Reveal Answers
Negative, because W = nRT ln(1/2) < 0.
Why it works: The final volume is smaller than the initial volume, so the logarithm is negative.
QUICK CHECK 3
Cycle orientation
Use the sign of the closed integral.
If the same P–V loop is traversed first clockwise and then counterclockwise, how are the two values of ∮P dV related?
Reveal Answers
They have equal magnitude and opposite sign.
Why it works: Reversing the path reverses the orientation of the line integral.
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
Use differentials to state the constraint
Isochoric, isobaric, isothermal, and adiabatic paths impose dV = 0, dP = 0, dT = 0, and δQ = 0.
KEY TAKEAWAY 2
Work is path dependent
Quasistatic work is W = ∫P(V)dV, so knowing only the endpoints is not enough unless the path is specified.
KEY TAKEAWAY 3
Integrals turn P–V geometry into energy
An isothermal ideal-gas path gives a logarithm, while a complete cycle gives Wcycle = ∮P dV, the signed enclosed area.
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