Adiabatic Process
adiabatic = no heat passage → Q = 0 (no heat exchange)
Process Summary
| Condition | Q = 0 |
| From First Law | ΔU = −W |
| PV relation | PVᵞ = constant |
| TV relation | TV^(γ−1) = constant |
| Work done by gas | W = nCv(T₁ − T₂) |
| PV graph shape | Steeper hyperbola than isothermal |
| Real example | Quick compression in cycle engine |
Thinking Step
Why does adiabatic expand cooler than isothermal? In isothermal: gas gets heat from surroundings to maintain T. In adiabatic: no heat input, so temperature drops as gas expands (uses internal energy for work).
Strategy Tip
On a PV diagram, adiabatic is ALWAYS steeper than isothermal through the same point. This is because slope of adiabatic = −γP/V, isothermal = −P/V, and γ > 1.
Isobaric Process
iso = same, baric = pressure → Pressure remains constant
Process Summary
| Condition | P = constant |
| From gas law | V/T = nR/P = constant |
| Work done by gas | W = PΔV = nRΔT |
| Heat added | Q = nCpΔT |
| Internal energy | ΔU = nCvΔT |
| PV graph shape | Horizontal line |
| Real example | Heating gas in balloon (P = atm) |
Exam Insight
Isobaric is the process where Cp is used (not Cv). Cp > Cv because extra energy goes into doing work PΔV. This is the physical meaning of Mayer's relation: Cp − Cv = R.
Strategy Tip
Quick check: In isobaric, W = nRΔT and ΔU = nCvΔT. So Q = ΔU + W = n(Cv+R)ΔT = nCpΔT. This confirms Cp = Cv + R without any derivation step.
Isochoric Process
iso = same, choric = volume → Volume remains constant
Process Summary
| Condition | V = constant (ΔV = 0) |
| Work done | W = 0 |
| From First Law | ΔU = Q |
| Heat added | Q = nCvΔT |
| From gas law | P/T = nR/V = constant |
| PV graph shape | Vertical line |
| Real example | Heating gas in rigid container |
Thinking Step
Isochoric is the simplest process for applying the First Law: W = 0, so ΔU = Q. Every joule of heat directly increases internal energy (and thus temperature). This is why Cv measures "pure" internal energy change per degree.
Common Mistake
Using Cp instead of Cv in isochoric process. At constant volume, use Q = nCvΔT. Using Cp gives a larger Q — this is wrong. The extra heat in Cp comes from work done against surroundings, which doesn't happen here.
📊 Master Comparison Table
This table alone can clear dozens of MCQs. Know it cold.
| Property | Isothermal | Adiabatic | Isobaric | Isochoric |
|---|---|---|---|---|
| Constant Quantity | T | Q (= 0) | P | V |
| Heat Q | W (= ΔU + W) | 0 | nCpΔT | nCvΔT |
| Work W | nRT ln(V₂/V₁) | nCv(T₁−T₂) | PΔV = nRΔT | 0 |
| ΔU | 0 | −W = nCvΔT | nCvΔT | Q = nCvΔT |
| PV Relation | PV = const | PVᵞ = const | V/T = const | P/T = const |
| PV Graph | Hyperbola | Steep hyperbola | Horizontal line | Vertical line |
| Slope of PV | −P/V | −γP/V | 0 | ∞ |
| Specific Heat | ∞ (any Q, ΔT=0) | 0 (no Q) | Cp | Cv |
♻️ Cyclic Process
A process where the system returns to its initial state after a series of steps.
Key Facts
ΔU_net = 0 for any cyclic process (returns to same state). W_net = area enclosed in PV diagram. Q_net = W_net (from First Law).
Thinking Step
Clockwise cycle = positive work (engine). Anticlockwise cycle = negative work (refrigerator). This is a 1-line answer to many JEE questions.
Carnot Cycle Steps on PV Diagram
A→B Isothermal Expansion
At T_H. Gas absorbs Q₁. Work = Q₁.
B→C Adiabatic Expansion
Q=0. T drops from T_H to T_C.
C→D Isothermal Compression
At T_C. Gas releases Q₂.
D→A Adiabatic Compression
Q=0. T rises back to T_H.