
For a real gas, the compressibility factor Z has different values at different temperature and pressures. Which of the following is not correct under the given condition?
a. Z < 1 at very low pressure
b. Z > 1 at high pressure
c. Z = 1 under all conditions
d. Z = 1 at intermediate pressure
Answer
219.9k+ views
Hint: The compressibility factor (Z) is a useful thermodynamic property which is used in modifying the ideal gas law to account for the behaviour of real gases. In simpler terms it can also be defined as the modifying factor for real gases.
Complete step by step answer: The compressibility factor (Z) is also known as the compression factor or the gas deviation factor. It is known to be a correction factor which is used to describe the deviation of a real gas from the ideal gas behaviour. Mathematically, it can be defined as the ratio of actual volume at a given pressure and temperature to the ideal volume under the same temperature and pressure conditions. For a real gas, the compressibility factor (Z) has different values for different sets of temperature and pressure conditions. The compressibility factor can be written as:
$ Z\quad =\quad \cfrac { PV }{ nRT }$
where Z is the compressibility factor, P is the pressure, V is the volume, n is the no. of moles, R is the gas constant and T is the temperature.
For real gases, when the pressure is high, the value of Z will be greater than 1 i.e., Z > 1.
And when the pressure is very low, the value of Z will be less than 1 i.e., Z < 1.
And at intermediate pressures, the value of Z will be equal to 1 i.e., Z = 1.
Therefore, the options (a), (b), and (d) are true. Thus, the only false option is option (c). Hence it is the correct answer.
Note: Since, an ideal gas is a hypothetical gas whose molecules occupy negligible space and does not have any interactions. For ideal gases, the gas law, $PV = nRT$ holds true. So, in that case, the compressibility factor, Z = 1 at any pressure and temperature conditions.
Complete step by step answer: The compressibility factor (Z) is also known as the compression factor or the gas deviation factor. It is known to be a correction factor which is used to describe the deviation of a real gas from the ideal gas behaviour. Mathematically, it can be defined as the ratio of actual volume at a given pressure and temperature to the ideal volume under the same temperature and pressure conditions. For a real gas, the compressibility factor (Z) has different values for different sets of temperature and pressure conditions. The compressibility factor can be written as:
$ Z\quad =\quad \cfrac { PV }{ nRT }$
where Z is the compressibility factor, P is the pressure, V is the volume, n is the no. of moles, R is the gas constant and T is the temperature.
For real gases, when the pressure is high, the value of Z will be greater than 1 i.e., Z > 1.
And when the pressure is very low, the value of Z will be less than 1 i.e., Z < 1.
And at intermediate pressures, the value of Z will be equal to 1 i.e., Z = 1.
Therefore, the options (a), (b), and (d) are true. Thus, the only false option is option (c). Hence it is the correct answer.
Note: Since, an ideal gas is a hypothetical gas whose molecules occupy negligible space and does not have any interactions. For ideal gases, the gas law, $PV = nRT$ holds true. So, in that case, the compressibility factor, Z = 1 at any pressure and temperature conditions.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

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

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

Other Pages
NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reaction

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

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

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

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

