MHT-CET Physics · Thermodynamics

The First Law of Thermodynamics

Heat Q given to a gas is shared between the rise in its internal energy ΔU and the work W it does on its surroundings, Q = ΔU + W; internal energy depends only on the state, so ΔU is the same along any path and zero round a cycle.

All Thermodynamics notes MHT-CET Physics strategy

Why this matters

37 PYQs, 4 of them HARD. Seventeen apply the first law directly — signs, cycles, work read from a graph, a moving container brought to rest. Seventeen ask how heat splits at constant pressure between internal energy and work, and three find the work done in an adiabatic change. Three cards.

Concept 1 of 3: Q = ΔU + W and its Signs

Count heat given TO the gas as positive and work done BY the gas as positive; then Q = ΔU + W. Work done by the gas is the area under its path on a P–V graph, positive when it expands. Internal energy is a property of the state, so two paths between the same states have the same ΔU even when Q and W differ, and a full cycle has ΔU = 0 so the net work equals the net heat. In an isothermal change of an ideal gas ΔU = 0 and Q = W; in an adiabatic change Q = 0 and ΔU = −W; at constant volume W = 0 and Q = ΔU. Kinetic energy of a container that stops suddenly becomes internal energy of the gas. Free expansion is too fast to pass through equilibrium states, so it cannot be drawn on a P–V diagram.

Definition

$Q = \Delta U + W$ : Q > 0 heat in, W > 0 work by the gas.
Cycle: $\Delta U = 0$ , $Q_{\text{net}} = W_{\text{net}}$ = area enclosed (clockwise on P–V positive).
Path independence: abc gives Q = 80, W = 35 ⇒ ΔU = 45; adc with Q = 65 ⇒ W = 20 cal.
Isothermal: ΔU = 0; adiabatic: ΔU = −W; isochoric: W = 0.
Compressed at constant P: work ON the gas $= P\Delta V$ adds to ΔU.
Container of molar mass M stopped: $\tfrac{1}{2}MV^2 = C_v\Delta T$ per mole (monoatomic ⇒ $\Delta T = \dfrac{MV^2}{3R}$ ).

First law

Q = \Delta U + W, \qquad W = \int P\,dV

Worked example

A gas at a constant pressure of 100 N/m² expands from 2 m³ to 5 m³ while absorbing 500 J. Change in internal energy?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q3Moderate

Example 1 · Thermodynamics · First Law, Internal Energy, and Work-Heat Relations

When a system is taken from state 'a' to state 'c' along a path abc, it is found that

Q = 80

cal and

W = 35

cal. Along path adc

Q = 65

cal, the work done

W

along path adc is

Getting the sign of work wrong

W is work done BY the gas. When the gas is compressed, W is negative and the work done ON it adds to its internal energy.

Reading the direction of a cycle

Round a cycle the net work is the enclosed area — positive if the path goes clockwise on a P–V graph, negative if anticlockwise. The same triangle gives +3PV or −3PV.

Drill 16 more on q = δu + w and its signs

Concept 2 of 3: How Heat Splits at Constant Pressure

At constant pressure the heat is Q = nC_pΔT, the rise in internal energy is ΔU = nC_vΔT, and the work is W = nRΔT. So the fractions are fixed by γ alone: ΔU/Q = 1/γ and W/Q = 1 − 1/γ. For a monoatomic gas (γ = 5/3) 60% goes to internal energy and 40% to work; for a diatomic gas (γ = 7/5) 5/7 and 2/7, so W : ΔU : Q = 2 : 5 : 7. The same relations give C_v = R/(γ − 1). Giving the same heat to a gas at constant pressure and at constant volume raises the temperature γ times more at constant volume.

Definition

$C_p - C_v = R$ , $C_v = \dfrac{R}{\gamma - 1}$ .
$\dfrac{\Delta U}{Q} = \dfrac{1}{\gamma}$ , $\dfrac{W}{Q} = \dfrac{\gamma - 1}{\gamma}$ , $\dfrac{Q}{W} = \dfrac{\gamma}{\gamma - 1}$ .
Monoatomic: 60% / 40%; diatomic: W : ΔU : Q = 2 : 5 : 7.
V → 2V at pressure P: $\Delta U = \dfrac{PV}{\gamma - 1}$ .
Same heat, piston free (A) or fixed (B): $dT_B = \gamma\,dT_A$ .

Isobaric split

\frac{\Delta U}{Q} = \frac{1}{\gamma}, \qquad \frac{W}{Q} = 1 - \frac{1}{\gamma}

Worked example

A diatomic gas (γ = 7/5) is given 700 J at constant pressure. How much goes into internal energy and how much into work?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q20Moderate

Example 2 · Thermodynamics · First Law, Internal Energy, and Work-Heat Relations

If heat energy

\Delta Q

is supplied to an ideal diatomic gas, the increase in internal energy is

\Delta U

and the amount of work done by the gas is

\Delta W

. The ratio

\Delta W : \Delta U : \Delta Q

Inverting the ratio

Q/W is γ/(γ − 1), and W/Q is its reciprocal (γ − 1)/γ. Both appear among the options; read which one the question asks for.

Using C_v for heat at constant pressure

Heat supplied at constant pressure is nC_pΔT; nC_vΔT is only the part that raises the internal energy. For 14 g of nitrogen warmed 48 °C that is 84R, not 60R.

Drill 16 more on how heat splits at constant pressure

Concept 3 of 3: Work in an Adiabatic Change

With no heat exchanged, all the work comes out of internal energy: W = −ΔU = nC_v(T₁ − T₂) = nR(T₁ − T₂)/(γ − 1). An expanding gas does positive work and cools. To find the final temperature first, use TV^(γ−1) = constant; for pressure against temperature, P ∝ T^(γ/(γ−1)).

Definition

$W = \dfrac{nR(T_1 - T_2)}{\gamma - 1} = \dfrac{P_1V_1 - P_2V_2}{\gamma - 1}$ .
1 mole, γ = 5/3, W = 6R ⇒ T falls by 4 K.
Volume doubled with γ = 3/2: $T_2 = \dfrac{T}{\sqrt{2}}$ , $W = RT(2 - \sqrt{2})$ .
$R = 0.4C_v$ ⇒ γ = 1.4 ⇒ $P \propto T^{7/2}$ .

Adiabatic work

W = \frac{nR\,(T_1 - T_2)}{\gamma - 1}

Worked example

Two moles of a diatomic gas (γ = 1.4) expand adiabatically and cool from 400 K to 300 K. Work done by the gas?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q30Moderate

Example 3 · Thermodynamics · First Law, Internal Energy, and Work-Heat Relations

One mole of an ideal gas at an initial temperature of

T

K does

6R

J of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is

5/3

, the final temperature of gas will be (

R = 8.31

J mol

^{-1}

^{-1}

)

Dividing by γ instead of γ − 1

The adiabatic work is nRΔT/(γ − 1), because it equals nC_vΔT and C_v = R/(γ − 1).

Drill 2 more on work in an adiabatic change

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Q = ΔU + W and its Signs
First law
$Q = \Delta U + W, \qquad W = \int P\,dV$
How Heat Splits at Constant Pressure
Isobaric split
$\frac{\Delta U}{Q} = \frac{1}{\gamma}, \qquad \frac{W}{Q} = 1 - \frac{1}{\gamma}$
Work in an Adiabatic Change
Adiabatic work
$W = \frac{nR\,(T_1 - T_2)}{\gamma - 1}$

Watch out for (5)

Getting the sign of work wrong
→ Q = ΔU + W and its Signs
Reading the direction of a cycle
→ Q = ΔU + W and its Signs
Inverting the ratio
→ How Heat Splits at Constant Pressure
Using C_v for heat at constant pressure
→ How Heat Splits at Constant Pressure
Dividing by γ instead of γ − 1
→ Work in an Adiabatic Change

Test yourself on Thermodynamics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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