Assertion : Compressibility factor (Z) for no | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The compressibility factor $(Z)$ is defined as $Z = \frac{PV}{nRT}$. For non-ideal gases,$Z$ can be greater than $1$ (when repulsive forces domina

Vedclass · Accepted Answer

If the Assertion is correct but Reason is incorrect.

Vedclass · Answer

(C) The compressibility factor $(Z)$ is defined as $Z = \frac{PV}{nRT}$. For non-ideal gases,$Z$ can be greater than $1$ (when repulsive forces dominate) or less than $1$ (when attractive forces dominate). Therefore,the Assertion is correct.Non-ideal gases do not always exert higher pressure than expected. Due to intermolecular attractive forces,the pressure exerted by a real gas is generally lower than the pressure predicted by the ideal gas law $(P_{real} < P_{ideal})$. Thus,the Reason is incorrect.

Assertion : Compressibility factor $(Z)$ for non-ideal gases can be greater than $1$.
Reason : Non-ideal gases always exert higher pressure than expected.

Similar Questions

Gas deviates from ideal gas nature because molecules

$1 \ mol$ of a real gas is kept at a high pressure of $100 \ bar$ at $300 \ K$. If the van der Waals constant $b$ is $0.005 \ L \ mol^{-1}$, what are the values of the compressibility factor $Z$ of the gas and the $\%$ deviation of volume from ideality?
$Z$ $\%$ Deviation

Explain the compressibility factor $(Z)$.

Which of the following indicates a plot of compressibility factor $(Z)$ $vs$ $p$ at room temperature for helium?

The deviation from the ideal gas equation $PV = nRT$ is maximum under which of the following conditions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion : Compressibility factor $(Z)$ for non-ideal gases can be greater than $1$.Reason : Non-ideal gases always exert higher pressure than expected.