Why does compressing a gas at low temperature make real gases deviate significantly from $Z = 1$?

At low temperatures, gas molecules possess reduced average kinetic energy ($E_k \propto T$), allowing short-range intermolecular attractive forces (quantified by parameter $a$ in the van der Waals equation) to dominate. At high pressures, molecules are compressed into small volumes, making their non-zero co-volume ($b$) significant relative to the total container volume. Both factors cause the compressibility factor $Z = \frac{P V_m}{R T}$ to diverge sharply from unity.

What is the key difference between Boyle temperature ($T_b$) and Critical temperature ($T_c$)?

Boyle temperature ($T_b = \frac{a}{R b}$) is the temperature at which a real gas obeys ideal gas laws over an appreciable pressure range because attractive and repulsive intermolecular forces effectively balance out ($Z \approx 1$). Critical temperature ($T_c = \frac{8 a}{27 R b}$), on the other hand, is the maximum temperature above which a gas cannot be liquefied regardless of the applied pressure. Notice that $T_c = \frac{8}{27} T_b \approx 0.296 T_b$.

How should I select the correct value and units of universal gas constant $R$ in numerical problems?

Always match $R$ to the units of pressure and volume: use $R = 8.314 \text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ for SI unit calculations (speed distributions $v_{\text{rms}}$, energy in Joules, pressure in $\text{Pa}$, volume in $\text{m}^3$), and $R = 0.0821 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ when working with pressure in $\text{atm}$ and volume in Litres.

States of Matter: Gases & Liquids Notes, Formulas, Traps & PYQs | ExamiFly

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Key Formulas & High-Yield Equations

Ideal Gas Equation of State

$P V = n R T$

Valid strictly for ideal gases under conditions of low pressure (

P \to 0

) and high temperature (

T \to \infty

), where intermolecular forces and point-particle volume are strictly negligible.

van der Waals Equation for Real Gases

$\left(P + \frac{a n^2}{V^2}\right)(V - n b) = n R T$

Applies to real gases; parameter

a

accounts for intermolecular attractive forces (units:

\text{atm}\cdot\text{L}^2\cdot\text{mol}^{-2}

), while

b

represents effective excluded co-volume (units:

\text{L}\cdot\text{mol}^{-1}

, where

b = 4 V_{\text{actual}}

Root Mean Square, Average, and Most Probable Speeds

$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{\rho}}$

M

must be substituted in

\text{kg}\cdot\text{mol}^{-1}

(SI units). The ratio of speeds follows

v_{\text{mp}} : v_{\text{avg}} : v_{\text{rms}} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} \approx 1 : 1.128 : 1.225

Compressibility Factor ($Z$)

$Z = \frac{P V_m}{R T} = \frac{V_{\text{real}}}{V_{\text{ideal}}}$

Measures deviation from ideality. For an ideal gas

Z = 1

. At low

P

Z < 1

(attractions dominate,

Z \approx 1 - \frac{a}{V_m R T}

). At high

P

Z > 1

(repulsions dominate,

Z \approx 1 + \frac{P b}{R T}

Graham's Law of Effusion

$\frac{r_1}{r_2} = \frac{P_1}{P_2} \sqrt{\frac{M_2}{M_1}}$

Valid for effusion of gases through a tiny orifice into vacuum/low-pressure zones; rate

r

can be expressed as

\frac{d V}{d t}

\frac{d n}{d t}

, or

\frac{d P}{d t}

Critical Constants ($T_c, P_c, V_c$)

$T_c = \frac{8 a}{27 R b}, \quad P_c = \frac{a}{27 b^2}, \quad V_c = 3 b$

Defines the critical state above which a gas cannot be liquefied, regardless of applied pressure. At the critical point,

Z_c = \frac{P_c V_c}{R T_c} = \frac{3}{8} = 0.375

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2024

Calculate the root mean square speed ( $v_{\text{rms}}$ ) of $\text{N}_2$ gas molecules at $300 \text{ K}$ . (Given: $R = 8.314 \text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ , Molar mass of $\text{N}_2 = 28 \text{ g}\cdot\text{mol}^{-1}$ )

Solution Locked

Step-by-Step Solution & Concept

Step 1

Identify the formula for root mean square speed $v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}$ and convert molar mass $M$ to SI units: $M = 28 \times 10^{-3} \text{ kg}\cdot\text{mol}^{-1}$ .

Step 2

Substitute the given parameters into the formula: $v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \times 300}{28 \times 10^{-3}}} = \sqrt{\frac{7482.6}{0.028}} = \sqrt{267235.71}$ .

Step 3

Evaluate the square root to obtain $v_{\text{rms}} \approx 516.95 \text{ m s}^{-1} \approx 517 \text{ m s}^{-1}$ .

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Question #2JEE Main 2023

For a real gas obeying the van der Waals equation, the critical pressure $P_c$ is $10 \text{ atm}$ and the critical temperature $T_c$ is $300 \text{ K}$ . Calculate the value of the van der Waals constant $b$ in $\text{L}\cdot\text{mol}^{-1}$ . (Given: $R = 0.0821 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ )

Solution Locked

Step-by-Step Solution & Concept

Step 1

Express critical parameters in terms of van der Waals constants: $T_c = \frac{8 a}{27 R b}$ and $P_c = \frac{a}{27 b^2}$ . Divide $T_c$ by $P_c$ to obtain $\frac{T_c}{P_c} = \frac{8 b}{R}$ .

Step 2

Rearrange the relationship to solve explicitly for constant $b$ : $b = \frac{R T_c}{8 P_c}$ .

Step 3

Substitute $R = 0.0821 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ , $T_c = 300 \text{ K}$ , and $P_c = 10 \text{ atm}$ : $b = \frac{0.0821 \times 300}{8 \times 10} = \frac{24.63}{80} \approx 0.0924 \text{ L}\cdot\text{mol}^{-1}$ .

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Mastering States of Matter: Gases and Liquids: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Ideal Gas Equation of State

van der Waals Equation for Real Gases

Root Mean Square, Average, and Most Probable Speeds

Compressibility Factor ($Z$)

Graham's Law of Effusion

Critical Constants ($T_c, P_c, V_c$)

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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