If the compressibility factor (Z) of a real g | 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 the ratio of the molar volume of a real gas $(V_m)$ to the molar volume of an ideal gas $(V_{ideal}

Vedclass · Accepted Answer

$23.52$

Vedclass · Answer

(C) The compressibility factor $(Z)$ is defined as the ratio of the molar volume of a real gas $(V_m)$ to the molar volume of an ideal gas $(V_{ideal})$ at the same temperature and pressure.$Z = \frac{V_m}{V_{ideal}}$At $STP$,the molar volume of an ideal gas $(V_{ideal})$ is $22.40 \ dm^3 \ mol^{-1}$.Given $Z = 1.05$,we can calculate the molar volume of the real gas $(V_m)$ as:$V_m = Z 	imes V_{ideal}$$V_m = 1.05 	imes 22.40 \ dm^3 \ mol^{-1}$$V_m = 23.52 \ dm^3 \ mol^{-1}$

If the compressibility factor $(Z)$ of a real gas is $1.05$ at $STP$,what is the molar volume of the real gas (in $dm^3$)?

Similar Questions

The compressibility factor of a real gas at high pressure is

In the van der Waals equation for gases,what does the constant $b$ represent?

Under which conditions will a real gas behave most like an ideal gas?

If the compressibility factor $(Z)$ is greater than unity,then which of the following is true for the molar volume $(V_m)$ of a gas at $NTP$?

One mole of a monoatomic real gas satisfies the equation $p(V-b)=RT$ where $b$ is a constant. The relationship of interatomic potential $V(r)$ and interatomic distance $r$ for the gas is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module