Home
Class 11
CHEMISTRY
Dominance of strong repulsive forces amo...

Dominance of strong repulsive forces among the molecules of the gas (`Z =` compressibility factor)

A

depends on Z and indicates that `Z = 1`

B

depends on Z and indicates that `Z gt 1`

C

depends on Z and indicates that `Z lt 1`

D

is independent of Z.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the dominance of strong repulsive forces among the molecules of a gas and its relationship with the compressibility factor (Z), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Compressibility Factor (Z)**: - The compressibility factor (Z) is defined as \( Z = \frac{PV}{RT} \), where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. It provides insight into how a real gas deviates from ideal gas behavior. 2. **Identifying the Role of Repulsive Forces**: - In the context of gases, strong repulsive forces dominate when the molecules are close together. This typically occurs at high pressures or low volumes, where the volume occupied by the gas molecules becomes significant compared to the volume of the container. 3. **Using the Van der Waals Equation**: - The Van der Waals equation for real gases is given by: \[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \] - Here, \( a \) accounts for attractive forces between molecules, and \( b \) accounts for the volume occupied by the gas molecules themselves. When strong repulsive forces dominate, we can consider \( a \) to be negligible (i.e., \( a = 0 \)). 4. **Simplifying the Equation**: - For a gas where repulsive forces are dominant, the equation simplifies to: \[ P(V - b) = RT \] - Rearranging this gives: \[ PV = P b + RT \] 5. **Relating to the Compressibility Factor**: - Dividing through by \( RT \): \[ \frac{PV}{RT} = \frac{Pb}{RT} + 1 \] - Thus, we can express Z as: \[ Z = 1 + \frac{Pb}{RT} \] 6. **Analyzing the Result**: - Since \( P \), \( b \), and \( R \) are all positive values, it follows that \( \frac{Pb}{RT} \) is also positive. Therefore, we conclude: \[ Z > 1 \] - This indicates that when strong repulsive forces dominate among gas molecules, the compressibility factor Z is greater than 1. 7. **Conclusion**: - Based on the analysis, the dominance of strong repulsive forces among the molecules of a gas indicates that \( Z > 1 \). Therefore, the correct answer to the question is option B: "depends on Z and indicates that Z is greater than 1."

To solve the question regarding the dominance of strong repulsive forces among the molecules of a gas and its relationship with the compressibility factor (Z), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Compressibility Factor (Z)**: - The compressibility factor (Z) is defined as \( Z = \frac{PV}{RT} \), where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. It provides insight into how a real gas deviates from ideal gas behavior. 2. **Identifying the Role of Repulsive Forces**: ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Define compressibility factor of a gas.

Under critical states of a gas for one mole of a gas, compressibility factor is :

The type of forces present among halogen molecules

In the above Question, near the point B, compressibility factor Z is about :

At critical temperature, pressure and volume. The compressibility factor (Z) is

Assertion : Liquids and solids are hard to compress . Reason : Magnitude of the repulsive forces between the molecules rises very rapidly as the distance separating the molecules decreases .

The compressibility factor of an ideal gas is

A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option above the gas and its compressibility factor (Z) is :

Number of N_(2) molecules present L vessel at NTP when compressibility factor is 1.2 is

The value of compressibility factor ( Z ) for an ideal gas is