Question

3 moles of a real gas occupies 8 L volume at 192 K and it has 2 atm pressure. If compressibility factor of the gas is z then find the value of ‘6z’. (nearest integer)

Answer 1

Given: 3 moles of a real gas occupies 8 L volume at 192 K and it has 2 atm pressure.

Compressibility factor of the gas is z  

To find:  the value of ‘6z’ (nearest integer)

Step by step Explanation:

Compressibility factor is (also called as gas deviation factor) defined as the ratio of volume of real gas at a given temperature and pressure to the Volume occupied by ideal gas at same temperature and pressure.

The formula for compressibility factor;

$Z = \frac{PV}{nRT}$

P is the pressure of the gas = 2 atm

V is the volume = 8 L

n is number of moles of gas = 3 moles

R is Universal Gas constant = 0.0821 L atm/ K mole

T is temperature of the gas = 192 K

$Z = \frac{2\times 8}{3\times 0.0821\times 192} \\Z = \frac{16}{47.2896} \\Z = 0.33$

The value of compressibility factor for the given gas is 0.33

Thus, the value of 6z would be;

$6z = 6\times 0.33\\6z = 1.98$

The value of 6z is 0.98

The nearest integer value of 6z would be 2.