
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
140.1k+ 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
Types of Solutions - Solution in Chemistry

Difference Between Crystalline and Amorphous Solid

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Number of sigma and pi bonds in C2 molecule isare A class 11 chemistry JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

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

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry
