Gaseous State Short Notes for JEE Main: Revision Notes PDF

Ideal gas equation and the gas laws

Ideal gas equation

9× asked, last 2022

pV = nRT,\qquad n = \frac{m}{M}

T

in K.

R

= 8.314 J mol⁻¹ K⁻¹ (Pa, m³) = 0.0821 L atm mol⁻¹ K⁻¹ = 0.083 L bar mol⁻¹ K⁻¹.

Not °C, mm Hg or mL in $pV=nRT$ . Use kelvin and the units of your $R$ (760 mm Hg = 1 atm, 1 bar = 10⁵ Pa, 1 L = 10⁻³ m³). 1 mol at 1 atm, 27 °C: $V=0.0821\times300=24.6$ L, not 2.2 L.
Quick method. Mixture of known total mass: get total $n$ from $pV=nRT$ , then solve $n_1+n_2=n$ and $n_1M_1+n_2M_2=$ mass. 3 mol of $\ce{H2}$ ( $M=2$ ) and He ( $M=4$ ) weighing 10 g: 2 mol He, i.e. 8 g.

Density and molar mass

5× asked, last 2022

pM = dRT,\qquad M = \frac{dRT}{p} = \frac{mRT}{pV}

d

in g L⁻¹ with

R

in L atm or L bar. At fixed

p

and

T

d\propto M

; for one gas at fixed

p

d\propto 1/T

d

(g L⁻¹)

p

(atm)O12123

M=16

M=32

Constant

T

p

d

through O, slope

RT/M

(lighter gas steeper)

p

(atm)

pV

(L atm)O123424.649.3300 K600 K

1 mol, constant

T

pV

p

is flat at

nRT

Ideal-gas plots: at constant $T$ , $p$ vs $d$ is a straight line through the origin (slope $RT/M$ ). For a fixed amount, $pV$ vs $p$ at constant $T$ is horizontal, and $V$ vs $T$ (in K) at constant $p$ is a line through the origin.
Quick method. Molar mass by weighing a bulb: gas mass = filled − empty mass; bulb volume = mass of liquid that fills it ÷ the liquid's density. Then $M=mRT/pV$ .

Boyle's law

3× asked, last 2023

p_1V_1 = p_2V_2\qquad (n,\ T\ \text{fixed})

p\propto 1/V

for a fixed amount at constant

T

. Balloon:

V\propto r^3

, so 4 times the radius gives 64 times less pressure.

V

(L)

p

(atm)

0.6V_1

V_1

p_1

1.67p_1

300 K600 K

p\propto1/V

+67\%

, not

+40\%

1 mol isotherms; the hotter one lies higher. Cutting

V

0.6V_1

raises

p

p_1/0.6

Bulbs joined at constant $T$ : $p_{\text{final}}=\sum p_iV_i/\sum V_i$ ; an evacuated bulb ( $p=0$ ) adds only volume. An ideal gas expanding into vacuum does no work, and its temperature does not change.
Not a 40% rise in $p$ for a 40% fall in $V$ . $p$ goes as $1/V$ , so $V_2=0.6V_1$ gives $p_2=p_1/0.6=1.67\,p_1$ , a 67% rise.

Charles's, Gay-Lussac's and Avogadro's laws

4× asked, last 2022

\frac{V_1}{T_1}=\frac{V_2}{T_2}\ (n,p),\qquad \frac{p_1}{T_1}=\frac{p_2}{T_2}\ (n,V),\qquad V\propto n\ (p,T)

T

in kelvin. A rigid tank or tyre (fixed

V

) follows Gay-Lussac. Equal volumes at the same

p

and

T

hold equal moles.

t

(°C)

V

(L)

-273.15

1001 atm2 atm

V

t

: every line meets

V=0

-273.15

°C (0 K)

T

(K)

p

O300600

p_0

2p_0

27 °C327 °C54 °C: only

1.09p_0

Fixed

V

p\propto T

in kelvin only, so doubling

p

from 27 °C needs 327 °C

Not °C in the ratio. Only kelvin temperature is proportional to $V$ or $p$ . Doubling the pressure of a sealed gas at 27 °C needs 600 K = 327 °C, not 54 °C. Convert back to °C only at the end.

Combined gas law

2× asked, last 2022

\frac{p_1V_1}{n_1T_1} = \frac{p_2V_2}{n_2T_2}

Ideal gas;

n_1=n_2

for a closed sample. Cancel what stays fixed.

p

V

units just match on both sides;

T

must be in K.

gas mass (kg)

p

(atm)O51024fullhalf the gas used

Rigid cylinder, fixed

T

p\propto n\propto

gas mass (total minus empty weight)

Quick method. Gas used from a rigid cylinder at fixed $T$ : $p\propto n\propto$ gas mass. Take gas mass = total − empty weight first: 10 kg of gas at 4 atm falling to 5 kg leaves 2 atm.

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Gaseous State revision notes

Ideal gas equation and the gas laws

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Mixtures of gases

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Ideal gas equation and the gas laws

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