FULL REVIEW
Full Review — Thermodynamic Processes and PV Diagrams — Algebra-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: Algebra-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
Connect each process name to its algebraic constraint.
State the defining condition for isochoric, isobaric, isothermal, and adiabatic processes.
Reveal Answers
Isochoric: ΔV = 0. Isobaric: ΔP = 0. Isothermal: ΔT = 0. Adiabatic: Q = 0.
Why it works: The constraint determines which relationships simplify.
ACTIVITY 2
Recall Activity 2
Verify the work units you will use.
Show why Pa·m³ has units of joules, and state the conversion from liters to cubic meters.
Reveal Answers
Pa·m³ = (N/m²)m³ = N·m = J. Also, 1 L = 10⁻³ m³.
Why it works: P–V area is an energy only when consistent SI units are used.
ACTIVITY 3
Recall Activity 3
Read a rectangular cycle.
A P–V rectangle is traversed clockwise. Which horizontal segment contributes positive work, which contributes negative work, and what is the sign of the net work?
Reveal Answers
The rightward high-pressure segment is positive, the leftward low-pressure segment is negative, and the net work is positive.
Why it works: The positive expansion area is larger than the negative compression 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
Process constraints determine useful equations
Classify the path before calculating. Constant volume makes boundary work zero. Constant pressure makes work a rectangle, PΔV. For an ideal gas at constant temperature, PV is constant. An adiabatic process has no heat transfer.
W = 0 (isochoric); W = PΔV (isobaric); P₁V₁ = P₂V₂ (isothermal ideal gas); Q = 0 (adiabatic)
Example: An ideal gas expanding isothermally from 2.0 L to 6.0 L drops from 2.4 × 10⁵ Pa to 8.0 × 10⁴ Pa.
Sensei note: A path can have changing pressure and changing volume; do not force a constant-pressure formula onto a curved path.
KEY CONCEPT 2
Calculate work from P–V geometry
Work by the gas is the signed area under the P–V path. For a constant-pressure path, W = PΔV. Vertical constant-volume segments have zero work. For simple piecewise paths, calculate the area of each segment and add with signs.
W = PΔV for constant pressure; work is signed P–V area
Example: At 1.5 × 10⁵ Pa, expansion from 2.0 L to 5.0 L gives W = 450 J.
Sensei note: Always convert L to m³ before using pressure in pascals.
KEY CONCEPT 3
Use enclosed area for simple cycles
A closed cycle returns the gas to its starting state. For a rectangular cycle, the magnitude of net work is ΔPΔV. The direction sets the sign: clockwise is positive work by the gas; counterclockwise is negative.
|Wnet| = ΔPΔV for a rectangular cycle; clockwise positive, counterclockwise negative
Example: A cycle spanning 1.0 × 10⁵ to 3.0 × 10⁵ Pa and 2.0 to 5.0 L has area 600 J; clockwise gives +600 J.
Sensei note: The enclosed area is net work, not the work on every individual segment.
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
Calculate one constant-pressure expansion completely.
A gas expands at 1.5 × 10⁵ Pa from 2.0 L to 5.0 L. Find W.
Reveal Answers
ΔV = 3.0 × 10⁻³ m³. W = PΔV = (1.5 × 10⁵)(3.0 × 10⁻³) = 450 J.
Why it works: A horizontal path has constant pressure, so its P–V area is a rectangle.
PRACTICE 2
Guided Problem
Compare two paths with the same volume change.
A gas expands by 3.0 L. Path A stays at 3.0 × 10⁵ Pa; Path B stays at 1.0 × 10⁵ Pa. Calculate both works and compare.
Reveal Answers
Path A: 900 J. Path B: 300 J. Path A does 600 J more work.
Why it works: For the same ΔV, W = PΔV is proportional to pressure.
PRACTICE 3
Independent Problem
Calculate a simple cycle from its dimensions.
A clockwise rectangular cycle spans pressures 1.0 × 10⁵ Pa to 3.0 × 10⁵ Pa and volumes 2.0 L to 5.0 L. Find Wnet.
Reveal Answers
ΔP = 2.0 × 10⁵ Pa and ΔV = 3.0 × 10⁻³ m³. The area is 600 J, so Wnet = +600 J.
Why it works: For a rectangle, the enclosed area is ΔPΔV, and clockwise sets a positive sign.
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
Constant-pressure work
Calculate in SI units.
At 2.0 × 10⁵ Pa, a gas expands from 1.0 L to 4.0 L. Find W.
Reveal Answers
W = (2.0 × 10⁵)(3.0 × 10⁻³) = 600 J.
Why it works: The volume increases by 3.0 L = 3.0 × 10⁻³ m³.
QUICK CHECK 2
Isothermal pressure
Use the ideal-gas isothermal relation.
P₁ = 3.0 × 10⁵ Pa, V₁ = 2.0 L, and V₂ = 5.0 L. Find P₂ for an isothermal process.
Reveal Answers
P₂ = P₁V₁/V₂ = 1.2 × 10⁵ Pa.
Why it works: At fixed temperature for a fixed amount of ideal gas, PV is constant.
QUICK CHECK 3
Cycle sign
Use both area and direction.
A rectangular P–V cycle is traversed counterclockwise. What is the sign of Wnet by the gas?
Reveal Answers
Negative.
Why it works: Compression occurs at higher pressure than expansion, so the signed enclosed area is negative.
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
Process constraints simplify the model
Classify the path first, then apply the relationship that matches the constraint.
KEY TAKEAWAY 2
P–V area has energy units
Use SI units so pressure times volume gives joules; horizontal work is PΔV and vertical work is zero.
KEY TAKEAWAY 3
Cycle area gives net work
For a rectangular cycle, the magnitude is ΔPΔV. Clockwise is positive work by the gas; counterclockwise is negative.
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