
At very high pressure, the compressibility factor of one mole of gas is given by:
(A) $1 + \dfrac{{pb}}{{RT}}$
(B) $1 - \dfrac{{pb}}{{RT}}$
(C) $1 - \dfrac{b}{{(VRT)}}$
(D) $\dfrac{{pb}}{{RT}}$
Answer
224.1k+ views
Hint: Compressibility factor is also known as gas deviation factor. This is a correction factor which describes the deviation of real gas from ideal gas.
Complete step by step answer:
The real gases obey the ideal gas equation $(PV = nRT)$ only if pressure is low or temperature is high.
Van der waals equation for one mole of Real gas is given by
$\Rightarrow \left[ {p + \dfrac{q}{{{V^2}}}} \right]\left[ {V - b} \right] = RT$ . . . . . (1)
And for $n$ moles of gas
$\Rightarrow \left[ {p + \dfrac{q}{{{V^2}}}} \right]\left[ {V - nb} \right] = RT$
Where ‘a’ and ‘b’ are constants and called Van der Waals constants. Their values depend upon the nature of gas.
At high pressure $\dfrac{a}{{{V^2}}}$ can be neglected.
$\therefore $equation (1) becomes
$\Rightarrow P[v - b] = RT$
$\Rightarrow PV - Pb = RT$. . . . . (2)
Dividing equation (2) by $RT$ we get
$\Rightarrow \dfrac{{PV}}{{RT}} - \dfrac{{Pb}}{{RT}} = 1$
$\Rightarrow \dfrac{{PV}}{{RT}} = 1 + \dfrac{{Pb}}{{RT}}$ . . . . . (3)
$\Rightarrow Z = \dfrac{{PV}}{{RT}}$ . . . . . (4)
$\Rightarrow Z = 1 + \dfrac{{Pb}}{{RT}}$
Where $Z$ = compressibility factor
Therefore, at high pressure compressibility factor is $1 + \dfrac{{Pb}}{{RT}}$
Therefore, by the above explanation, the correct option is [A] $1 + \dfrac{{Pb}}{{RT}}$
Additional information:
At high pressure, compressibility factor is greater than $1$.
As $P$ is increased (at constant T). The factor $\dfrac{{Pb}}{{RT}}$ increases. This explains how compressibility increases continuously with pressure.
Van der Waals constants $a$ and $b$ have significance of attractive force among molecules of gas and value of $b$ is a measure of the effective size of gas molecules.
Note: A gas which obeys the ideal gas equation $PV = nRT$ under all conditions of temperature and pressure is called ideal gas. Concept of ideal gas is only theoretical when pressure is low and temperature is high. They obey the gas law. Such gases are known as Real gases. All gases are Real gases.
Complete step by step answer:
The real gases obey the ideal gas equation $(PV = nRT)$ only if pressure is low or temperature is high.
Van der waals equation for one mole of Real gas is given by
$\Rightarrow \left[ {p + \dfrac{q}{{{V^2}}}} \right]\left[ {V - b} \right] = RT$ . . . . . (1)
And for $n$ moles of gas
$\Rightarrow \left[ {p + \dfrac{q}{{{V^2}}}} \right]\left[ {V - nb} \right] = RT$
Where ‘a’ and ‘b’ are constants and called Van der Waals constants. Their values depend upon the nature of gas.
At high pressure $\dfrac{a}{{{V^2}}}$ can be neglected.
$\therefore $equation (1) becomes
$\Rightarrow P[v - b] = RT$
$\Rightarrow PV - Pb = RT$. . . . . (2)
Dividing equation (2) by $RT$ we get
$\Rightarrow \dfrac{{PV}}{{RT}} - \dfrac{{Pb}}{{RT}} = 1$
$\Rightarrow \dfrac{{PV}}{{RT}} = 1 + \dfrac{{Pb}}{{RT}}$ . . . . . (3)
$\Rightarrow Z = \dfrac{{PV}}{{RT}}$ . . . . . (4)
$\Rightarrow Z = 1 + \dfrac{{Pb}}{{RT}}$
Where $Z$ = compressibility factor
Therefore, at high pressure compressibility factor is $1 + \dfrac{{Pb}}{{RT}}$
Therefore, by the above explanation, the correct option is [A] $1 + \dfrac{{Pb}}{{RT}}$
Additional information:
At high pressure, compressibility factor is greater than $1$.
As $P$ is increased (at constant T). The factor $\dfrac{{Pb}}{{RT}}$ increases. This explains how compressibility increases continuously with pressure.
Van der Waals constants $a$ and $b$ have significance of attractive force among molecules of gas and value of $b$ is a measure of the effective size of gas molecules.
Note: A gas which obeys the ideal gas equation $PV = nRT$ under all conditions of temperature and pressure is called ideal gas. Concept of ideal gas is only theoretical when pressure is low and temperature is high. They obey the gas law. Such gases are known as Real gases. All gases are Real gases.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip Releasing Today, Application Form Closed, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

