Thermodynamics

ThermodynamicsNEET Physics · Class 11 · NCERT Chapter 11

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Thermodynamics. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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First law solver (Q = ΔU + W)

Cp / Cv / γ for ideal gas (mono / di / poly)

PV diagram process explorer (4 processes)

Isothermal vs adiabatic (same start)

Cyclic process — net work as enclosed area

Carnot heat engine efficiency

Refrigerator and heat pump COP

First law and gas specific heats

The energy bookkeeping of a thermodynamic system, plus how Cp, Cv and gamma come from degrees of freedom.

First law

First law of thermodynamics solver

Pick the unknown and the calculator returns it from Q = ΔU + W. NEET asks this in every form: signs, units and direction matter.

Pick which quantity to solve for. Enter the other two. Sign convention: Q positive when heat enters, W positive when gas expands.

Q (J)

ΔU (J)

W (J)

Heat added to system Q

500.0 J

Q = Δ U + W

Try this

Isothermal expansion: ΔU = 0, so Q = W. All heat goes into work.
Adiabatic compression: Q = 0, so ΔU = −W. All work goes into raising internal energy.
Isochoric heating: W = 0 (no volume change), so Q = ΔU. All heat raises temperature.
Free expansion into vacuum: Q = 0, W = 0, so ΔU = 0 (T stays the same for an ideal gas).

Cp / Cv / γ

Specific heats of an ideal gas

Cv from degrees of freedom, Cp from Mayer's relation, gamma from their ratio. Pick a gas type and compare.

Degrees of freedom f set Cv. Mayer's relation gives Cp. Their ratio is gamma.

Examples: H₂, O₂, N₂

Degrees of freedom (f)

Cv = (f/2) R

20.79 J/(mol·K)

Cp = Cv + R

29.10 J/(mol·K)

γ = Cp / Cv

1.400

C_{V} = \frac{f}{2} R, C_{P} = C_{V} + R, γ = \frac{C _{P}}{C _{V}} = 1 + \frac{2}{f}

Try this

Monoatomic gases (helium, argon): f = 3, gamma = 5/3 ≈ 1.667.
Diatomic gases at room temp (oxygen, nitrogen): f = 5, gamma = 7/5 = 1.4.
Higher gamma means a steeper adiabat: a small volume change makes a big pressure change.
Mayer relation Cp − Cv = R is exact only for ideal gases.

PV processes and work

The four NEET processes side by side, plus a direct isothermal vs adiabatic comparison and cyclic-process net work.

PV processes

PV diagram process explorer

Switch between the four NEET processes. The shaded area is the work done by the gas, read it directly from the chart.

P₁: 4.0 atm

V₁: 1.00 L

V₂: 3.00 L

Final pressure P₂

1.33 atm

Work done by gas (area under curve)

W = 4.39 L·atm

(1 L·atm ≈ 101.3 J)

Isothermal (T constant)

P V = const, W = n R T ln \frac{V _{2}}{V _{1}}

Try this

Switch to adiabatic and watch the curve drop more steeply than the isothermal one.
Isochoric (constant V): the curve becomes a vertical line, so no area, no work.
Isobaric: rectangle. Area = P × ΔV.
Isothermal expansion at high T does more work than at low T (work scales with nRT).

Iso vs adiabatic

Isothermal vs adiabatic, same start, different paths

Compare the two most-tested processes side by side. The adiabat is always steeper because the gas cools as it expands.

Same start point, same final volume, but different processes. The adiabat is steeper than the isotherm because temperature also drops (during expansion). For the same expansion the adiabat does less work.

P₁: 4.0 atm

V₁: 1.00 L

V₂: 2.50 L

γ (adiabat): 1.40

Isothermal P₂

1.60 atm

W = 3.67 L·atm

Adiabatic P₂

1.11 atm

W = 3.07 L·atm

● Isothermal (PV = const)

● Adiabatic (PVᵞ = const)

Iso: P V = P_{1} V_{1} Adi: P V^{γ} = P_{1} V_{1}^{γ}

Try this

For expansion, isothermal does MORE work than adiabatic.
For compression (drag V₂ below V₁ if you allow it), adiabatic needs MORE work because temperature rises.
Slope ratio: adiabatic slope / isothermal slope = γ. Steeper adiabat for any γ > 1.
Helium (γ = 5/3) has the steepest adiabat. CO₂ (γ ≈ 1.3) is closer to the isotherm.

Cyclic process

Net work in a cyclic process

Net work in one cycle equals the enclosed area on a PV diagram. Direction sets the sign: clockwise is an engine, counter-clockwise is a refrigerator.

A rectangular cycle on the PV diagram. Net work in one cycle = enclosed area. Direction matters: clockwise = engine (work done by gas, positive); counter-clockwise = refrigerator (work done on gas, negative).

P₁ (low): 1.0 atm

P₂ (high): 4.0 atm

V₁ (small): 1.0 L

V₂ (large): 4.0 L

Net work in one cycle

9.00 L·atm

ΔU = 0 over a complete cycle (state returns to start). So Q_net = W_net.

Engine: gas does net positive work each cycle

W_{net} = \oint P d V = enclosed area

Try this

Make the rectangle larger by spreading P and V; net work scales with the enclosed area.
Switch to counter-clockwise: same area, but the gas now needs work input each cycle.
For ANY closed cycle (even non-rectangular), net work = enclosed area on the PV diagram.
Over one full cycle ΔU = 0 (state returns), so net heat absorbed = net work done.

Heat engines and refrigerators

Carnot efficiency between two reservoirs, and the same machine run in reverse as a refrigerator or heat pump.

Carnot engine

Carnot heat engine, efficiency limit

The maximum efficiency any engine can achieve between two reservoirs. Real engines fall below this limit.

Carnot is the most efficient possible engine between two reservoirs. Real engines fall below this limit.

Hot reservoir T_h: 600 K

Cold reservoir T_c: 300 K

Heat absorbed Q_h: 1000 J

Carnot efficiency

η = 50.00%

η = 1 - \frac{T _{c}}{T _{h}}

Useful work W

500 J

Wasted heat Q_c

500 J

Try this

Increase T_h or decrease T_c, and efficiency rises. Both must be in kelvin.
Set T_c = 0 K (impossible in practice): efficiency would be 100%. Third law forbids this.
No real engine between 600 K and 300 K can beat 50% efficiency.
For T_h = 400 K, T_c = 300 K: η = 25%. The temperature ratio matters more than the absolute heat input.

Refrigerator COP

Refrigerator and heat pump COP

A fridge moves heat from cold to hot using work. COP measures how much heat is moved per unit work.

A refrigerator pumps heat from cold to hot. You pay work W and you get Q_c removed from the inside. The same machine used in reverse becomes a heat pump.

Room T_h: 310 K

Cold space T_c: 273 K

Work input W per cycle: 200 J

COP refrigerator

7.38

= T_c / (T_h − T_c)

COP heat pump

8.38

= 1 + COP_fridge

Heat removed Q_c

1476 J

Heat dumped Q_h

1676 J

COP_{fridge} = \frac{Q _{c}}{W} = \frac{T _{c}}{T _{h} - T _{c}}

Try this

A typical home fridge: T_c ≈ 273 K, T_h ≈ 310 K. COP ≈ 7.4 in the ideal case; real fridges hit around 2 to 4.
As T_c falls (deeper freezer), COP drops fast, since moving heat from very cold places needs much more work.
Heat pump = same device used to heat your home. COP_HP > 1 means more heat delivered to room than work used.
Q_h is always > Q_c, and the extra heat equals the work you put in (energy conservation).

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