MHT-CET Chemistry · Some Basic Concepts of Chemistry

Gas Laws and the Ideal Gas Equation

Four simple gas laws (Boyle, Charles, Gay-Lussac, combined) each fix one variable and relate the rest; the ideal gas equation PV = nRT ties pressure, volume, moles and temperature together in a single formula.

All Some Basic Concepts of Chemistry notes MHT-CET Chemistry strategy

Why this matters

This subtopic is a reliable scoring block in MHT-CET Chemistry — most of its PYQs are direct one-step plug-ins (compress a gas, cool a balloon, find the temperature from PV = nRT). The recurring traps are always the same: temperature must be in kelvin, R's units must match the pressure's units, and for equal masses the lightest gas exerts the highest pressure. Learn the four proportionalities plus PV = nRT cold and you can attempt every question here on sight.

Concept 1 of 6

Boyle's law — pressure and volume

Intuition

Squeeze a gas into half the space and it pushes back with twice the pressure. At constant temperature, pressure and volume trade off inversely — one goes up exactly as much as the other goes down. Their product P times V stays constant, so a P-versus-V graph is a hyperbola.

Definition

Boyle's law (constant temperature, fixed mass):

Pressure is inversely proportional to volume: $P \propto \dfrac{1}{V}$ .
Equivalently $PV = \text{constant}$ , so $P_1 V_1 = P_2 V_2$ .
The P vs V plot is a rectangular hyperbola; a plot of $PV$ vs $P$ is a horizontal straight line.
Units of P and V may be anything as long as they are the same on both sides (atm with atm, mL with mL) — no unit conversion needed.

Boyle's law

P_1 V_1 = P_2 V_2 \qquad (T,\, n \text{ constant})

P_1, V_1initial pressure and volume
P_2, V_2final pressure and volume

Worked example

A gas occupies 500 mL at 1 atm. It is compressed to 200 mL at the same temperature. What is the new pressure?

Temperature is constant, so use Boyle's law $P_1 V_1 = P_2 V_2$ .
$P_2 = \dfrac{P_1 V_1}{V_2} = \dfrac{1 \times 500}{200}$ .

Answer:

P_2 = 2.5

atm.

Practice this conceptself-check · 4 quick reps

Try it yourself

A gas occupies 11.2 dm3 at 105 kPa. What volume does it occupy if the pressure is increased to 210 kPa at constant temperature?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Volume of a gas at 1 atm is 25 mL. Volume at 1.25 atm (same T)?
2.
Gas at 2 atm occupies 6 L. Pressure if squeezed to 3 L?
3.
What shape is the P vs V graph for Boyle's law?
4.
1 dm3 of gas at NTP (1.013 x 10^5 Pa). Volume at 1.032 x 10^5 Pa?

From the bank · past-year question

Example 1States of MatterEASY

\text{N}_2

gas is compressed at 2 atmosphere from 9.0 L to 3.0 L at 300 K, find the final pressure at same temperature?

[Q74 · 11th May Shift 2 · 2024]

No unit conversion inside Boyle's law

Because P and V appear on both sides, they only need to be consistent, not SI. Keep atm with atm and mL with mL — converting to Pa or m3 wastes time and invites arithmetic slips.

Boyle's law is P vs V — not PV vs P

In a 'which graph explains Boyle's law?' question, the answer is the P-vs-V hyperbola. A straight horizontal line is the

PV

-vs-

P

plot (also a consequence, but not the direct Boyle graph the bank usually wants).

Concept 2 of 6

Charles' law — volume and temperature

Intuition

Heat a gas at constant pressure and it expands; cool it and it shrinks. Volume rises in lock-step with the ABSOLUTE (kelvin) temperature — double the kelvin temperature and you double the volume. This is why a hot-air balloon rises: warmed air takes up more volume, so it is less dense.

Definition

Charles' law (constant pressure, fixed mass):

Volume is directly proportional to absolute temperature: $V \propto T$ .
Equivalently $\dfrac{V}{T} = \text{constant}$ , so $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ .
T must be in kelvin: $T(\text{K}) = t(^{\circ}\text{C}) + 273$ .
Extrapolating V to zero gives $-273.15\,^{\circ}\text{C}$ = absolute zero (0 K), the lowest possible temperature.

Charles' law

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \qquad (P,\, n \text{ constant})

V_1, V_2initial and final volume
T_1, T_2initial and final absolute temperature (K)

Worked example

A hot-air balloon holds 2000 dm3 of air at 99 degrees C. What volume does the air occupy when it cools to 80 degrees C at constant pressure?

Convert to kelvin: $T_1 = 99 + 273 = 372\ \text{K}$ , $T_2 = 80 + 273 = 353\ \text{K}$ .
Charles' law: $V_2 = \dfrac{V_1 T_2}{T_1} = \dfrac{2000 \times 353}{372}$ .

Answer:

V_2 \approx 1897.8\ \text{dm}^3

(cooling shrinks the gas).

Practice this conceptself-check · 4 quick reps

Try it yourself

At 0 degrees C a gas occupies 22.4 L. To what temperature (in kelvin) must it be heated at constant pressure to occupy 224 L?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Gas heated 273 K to 373 K at 1 atm, initial volume 10 L. Final volume?
2.
Absolute zero in degrees Celsius?
3.
Double the kelvin temperature of a gas at constant P. What happens to its volume?
4.
Convert 27 degrees C to kelvin.

From the bank · past-year question

Example 2States of MatterEASY

A gas is heated from 273 K to 373 K at 1 atm pressure. If the initial volume of the gas is 10 L, its final volume would be

[Q57 · May Shift 1 · 2021]

Kelvin, always — never Celsius in the ratio

V/T

is constant only for the absolute temperature. Plugging in

99

and

80

(degrees C) instead of

372

and

353

K gives a badly wrong answer. Convert first:

T(\text{K}) = t(^{\circ}\text{C}) + 273

Absolute zero is negative

0\ \text{K} = -273.15\,^{\circ}\text{C}

, not

+273.15

. All molecular motion ceases here, and it is the temperature at which a gas's extrapolated volume would reach zero.

Concept 3 of 6

Gay-Lussac's law — pressure and temperature

Intuition

Seal a gas in a rigid container (fixed volume) and heat it: the molecules hit the walls harder and more often, so the pressure climbs. At constant volume, pressure rises directly with the absolute temperature. This is why an aerosol can warns against heating — trapped gas pressure grows with temperature.

Definition

Gay-Lussac's (pressure) law (constant volume, fixed mass):

Pressure is directly proportional to absolute temperature: $P \propto T$ .
Equivalently $\dfrac{P}{T} = \text{constant}$ , so $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ .
T must be in kelvin, exactly as in Charles' law.
Do not confuse the three simple laws: Boyle fixes T ( $PV=$ const), Charles fixes P ( $V/T=$ const), Gay-Lussac fixes V ( $P/T=$ const).

Gay-Lussac's law

\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \qquad (V,\, n \text{ constant})

P_1, P_2initial and final pressure
T_1, T_2initial and final absolute temperature (K)

Worked example

A rigid gas cylinder reads 2 atm at 300 K. What is the pressure when it is heated to 450 K (volume unchanged)?

Volume is fixed, so use Gay-Lussac's law $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ .
$P_2 = \dfrac{P_1 T_2}{T_1} = \dfrac{2 \times 450}{300}$ .

Answer:

P_2 = 3

atm.

Practice this conceptself-check · 3 quick reps

Try it yourself

State the mathematical form of Gay-Lussac's law and name the quantity held constant.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Which quantity is held constant in Gay-Lussac's law?
2.
Gas at 1 atm and 250 K. Pressure at 500 K (fixed volume)?
3.
Write Gay-Lussac's law as a ratio equation.

From the bank · past-year question

Example 3States of MatterEASY

Which from following is true according to Gay-Lussac's law?

[Shift || · 2025]

Do not mix up the three simple laws

V/T = \text{const}

is Charles' law, and

PV = \text{const}

is Boyle's law. Gay-Lussac's law is

P/T = \text{const}

at constant volume — the odd one out that fixes V, not T or P.

Absolute temperature here too

P/T

is constant only with T in kelvin. Convert any Celsius temperature before forming the ratio.

Concept 4 of 6

Combined gas law

Intuition

When pressure, volume AND temperature all change at once, no single simple law works. Merge Boyle and Charles into one relation: the quantity PV/T stays constant for a fixed mass of gas. It reduces to any one simple law when you hold the third variable fixed.

Definition

Combined gas law (fixed mass):

$\dfrac{PV}{T} = \text{constant}$ , so $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ .
Set $T_1 = T_2$ and it collapses to Boyle's law; set $P_1 = P_2$ and it becomes Charles' law; set $V_1 = V_2$ and it becomes Gay-Lussac's law.
T must be in kelvin. P and V just need to be consistent on both sides.

Combined gas law

\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}

P_1, V_1, T_1initial pressure, volume, absolute temperature
P_2, V_2, T_2final pressure, volume, absolute temperature

Worked example

A gas occupies 2 L at 300 K and 1 atm. What volume does it occupy at 600 K and 2 atm?

All three change, so use $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ .
$V_2 = \dfrac{P_1 V_1 T_2}{T_1 P_2} = \dfrac{1 \times 2 \times 600}{300 \times 2}$ .

Answer:

V_2 = 2

L (the doubled temperature and doubled pressure cancel out).

Practice this conceptself-check · 3 quick reps

Try it yourself

Which single equation expresses the combined relationship of Boyle's and Charles' laws?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Combined gas law reduces to which law when temperature is constant?
2.
It reduces to which law when pressure is constant?
3.
Gas: 1 atm, 1 L, 273 K. Volume at 2 atm and 546 K?

From the bank · past-year question

Example 4States of MatterEASY

Which of the following equations gives combined relationship of Boyle's law and Charle's law?

[Q96 · 16th May Shift 1 · 2023]

T on the DENOMINATOR, in kelvin

The temperature sits under PV:

PV/T

. A common slip is writing

PVT = \text{const}

or leaving T in Celsius. Keep the form

\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}

with kelvin temperatures.

Concept 5 of 6

Ideal gas equation, PV = nRT

Intuition

The ideal gas equation is the master formula: it bundles all the simple laws and adds the amount of gas (moles) explicitly. Given any three of P, V, n, T you can find the fourth — as long as R's units match the pressure's units. You can also swap n for mass/molar mass, which lets you weigh a gas.

Definition

The ideal gas equation:

$PV = nRT$ , and since $n = \dfrac{m}{M}$ : $PV = \dfrac{m}{M}RT$ , which rearranges to $PM = \rho RT$ ( $\rho$ = density).
The value of R decides the units:
$R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ — use with SI units: P in Pa (N m $^{-2}$ ), V in m $^3$ .
$R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$ — use with P in atm, V in litres.
T is always in kelvin. STP/NTP means $0\,^{\circ}\text{C}$ (273 K) or $25\,^{\circ}\text{C}$ (298 K) and 1 atm; convert $\text{dm}^3 \to \text{m}^3$ by $\div 10^3$ when using the SI R.

Ideal gas equation

PV = nRT = \dfrac{m}{M}RT

Ppressure (Pa with R = 8.314; atm with R = 0.0821)
Vvolume (m^3 with R = 8.314; L with R = 0.0821)
nnumber of moles, = m/M
Rgas constant (8.314 J K^-1 mol^-1 or 0.0821 L atm K^-1 mol^-1)
Tabsolute temperature (K)

Worked example

What volume is occupied by 3.2 g of oxygen gas at 2 atm and 273 K? (O2 = 32 g/mol, R = 0.0821 L atm K^-1 mol^-1)

Moles $n = \dfrac{m}{M} = \dfrac{3.2}{32} = 0.1$ mol.
P is in atm and the answer is wanted in litres, so use $R = 0.0821$ and $V = \dfrac{nRT}{P}$ .
$V = \dfrac{0.1 \times 0.0821 \times 273}{2} = \dfrac{2.241}{2} = 1.12\ \text{L}$ .

Answer:

1.12

litres.

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the temperature (in degrees C) of 2 mol of an ideal gas that occupies 20 dm3 at 4.926 atm. (R = 0.0821 dm3 atm K^-1 mol^-1)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Value of R in L atm K^-1 mol^-1?
2.
Value of R in SI units (J K^-1 mol^-1)?
3.
Rearrange PV = nRT to make P the subject.
4.
Convert 68 mL to m^3 for use with the SI value of R.

From the bank · past-year question

Example 5States of MatterEASY

How many moles of dioxygen are present in

8.314\times10^{-3}\ \text{m}^3

of it at 318 K having pressure

3.18\times10^5\ \text{Nm}^{-2}

? [

R=8.314\ \text{J K}^{-1}\ \text{mol}^{-1}

]

[Q76 · 3rd May 2nd Shift · 2023]

Match R's units to the pressure and volume

Using

R = 0.0821

with pressure in Pa, or

R = 8.314

with volume in litres, gives an answer off by orders of magnitude. Decide the R value FIRST from the units given, then convert everything to match (litres <-> m

^3

, atm <-> Pa).

Watch a printed exponent typo

For

3.4

mol in

68\ \text{mL} = 68 \times 10^{-6}\ \text{m}^3

at 300 K,

P = \dfrac{nRT}{V} = \dfrac{3.4 \times 8.314 \times 300}{68 \times 10^{-6}} = 1.247 \times 10^{8}\ \text{Pa} = 1.247 \times 10^{5}\ \text{kPa}

. Some printed papers mis-type this as

1.247 \times 10^{2}

kPa — trust your derivation; the mantissa 1.247 is what matches.

Concept 6 of 6

Equal masses in equal volumes — lightest gas, highest pressure

Intuition

Take the same mass of several gases in identical containers at the same temperature. The lighter the gas, the more molecules that mass contains, so more molecules hammer the walls — higher pressure. Pressure ends up inversely proportional to molar mass, so hydrogen always wins.

Definition

Same mass, same V and T:

From $PV = nRT$ with V and T fixed: $P \propto n$ .
With equal mass, $n = \dfrac{m}{M} \propto \dfrac{1}{M}$ , hence $P \propto \dfrac{1}{M}$ .
The gas with the lowest molar mass has the most moles and exerts the highest pressure.
Related quantity: vapour density $= \dfrac{M}{2}$ (molar mass relative to $\text{H}_2$ ), so a lighter gas also has a smaller vapour density.

Pressure vs molar mass (equal mass, V, T)

P \propto \dfrac{1}{M} \qquad \left(n = \dfrac{m}{M}\right)

Ppressure exerted by the gas
Mmolar mass of the gas
mmass of gas (same for all in the comparison)

Worked example

Four identical flasks each hold 4 g of a different gas at the same temperature: H2, N2, O2, Cl2. Which gas exerts the greatest pressure?

Equal mass, equal V, equal T, so $P \propto n \propto \dfrac{1}{M}$ .
Molar masses: $\text{H}_2 = 2 < \text{N}_2 = 28 < \text{O}_2 = 32 < \text{Cl}_2 = 71$ .
The smallest M gives the largest number of moles and the highest pressure.

Answer:

\text{H}_2

exerts the greatest pressure (lowest molar mass).

Practice this conceptself-check · 3 quick reps

Try it yourself

What is the vapour density of O2 gas? (molar mass of O2 = 32)

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Equal masses of H2 and O2 in identical flasks at the same T. Which has more moles?
2.
Same mass of gases, same V and T. Highest pressure goes to the gas with which molar mass?
3.
Vapour density of a gas with molar mass 44 (e.g. CO2)?

From the bank · past-year question

Example 6States of MatterMODERATE

Four vessels of same volume consist equal masses of four gases

H_{2},Cl_{2},{\text{ }N}_{2}

, and

O_{2}

separately at same temperature. The pressure exerted by the gas is maximum for

[Q90 · 19 April Shift I · 2025]

Equal MASS, not equal moles

This trick only works because the masses are equal — then

P \propto 1/M

. If instead the moles were equal, all four gases would exert the SAME pressure (same n, V, T). Read whether the question fixes mass or moles.

Lightest gas = highest pressure

It is easy to guess the heaviest gas (

\text{Cl}_2

) exerts the most pressure — the opposite is true. Fewer moles per gram means

\text{Cl}_2

gives the lowest pressure;

\text{H}_2

gives the highest.

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 (6)

Boyle's law — pressure and volume
Boyle's law
$P_1 V_1 = P_2 V_2 \qquad (T,\, n \text{ constant})$
Charles' law — volume and temperature
Charles' law
$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \qquad (P,\, n \text{ constant})$
Gay-Lussac's law — pressure and temperature
Gay-Lussac's law
$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \qquad (V,\, n \text{ constant})$
Combined gas law
Combined gas law
$\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$
Ideal gas equation, PV = nRT
Ideal gas equation
$PV = nRT = \dfrac{m}{M}RT$
Equal masses in equal volumes — lightest gas, highest pressure
Pressure vs molar mass (equal mass, V, T)
$P \propto \dfrac{1}{M} \qquad \left(n = \dfrac{m}{M}\right)$

Watch out for (11)

No unit conversion inside Boyle's law
→ Boyle's law — pressure and volume
Boyle's law is P vs V — not PV vs P
→ Boyle's law — pressure and volume
Kelvin, always — never Celsius in the ratio
→ Charles' law — volume and temperature
Absolute zero is negative
→ Charles' law — volume and temperature
Do not mix up the three simple laws
→ Gay-Lussac's law — pressure and temperature
Absolute temperature here too
→ Gay-Lussac's law — pressure and temperature
T on the DENOMINATOR, in kelvin
→ Combined gas law
Match R's units to the pressure and volume
→ Ideal gas equation, PV = nRT
Watch a printed exponent typo
→ Ideal gas equation, PV = nRT
Equal MASS, not equal moles
→ Equal masses in equal volumes — lightest gas, highest pressure
Lightest gas = highest pressure
→ Equal masses in equal volumes — lightest gas, highest pressure

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