Question

The density of steam at $100^oC$ and $1×10^5$ Pa is $0.6kg m^-3$. The compressibility factor for steam is:
A.1
B.0.967
C.1.03
D.0.867

Answer 1

Hint: The compressibility factor is defined as that factor which describes the deviation of a real gas from ideal gas behaviour. The compressibility factor is denoted by “Z”.

Complete step by step answer:
The formula used in this numerical is as given:
$Z = \dfrac{{PV}}{{nRT}}$ Where n is the number of moles and P is the pressure.
And R is a gas constant and T is the temperature.

As per the question, we have:
\[{\text{Pressure(P)}} = 1 \times {10^5}Pa\]
$
  {\text{Temperature(T)}} = 100^\circ C \\
  {\text{Density}} = 0.6{\text{ kg/}}{{\text{m}}^{{\text{ - 3}}}} \\
$
As we know the value of gas constant, $R = 8.314{m^3}Pa{K^{ - 1}}mo{l^{ - 1}}$ and number of moles (n) is given by the formula:
$n = \dfrac{{given{\text{ }}mass{\text{ }}(w)}}{{molar{\text{ }}mass{\text{ }}(m)}}$
By putting, the formula of compressibility factor, we get:
$Z = \dfrac{{PVm}}{{wRT}}$
Since, density is the mass per unit volume and so the above formula becomes:
$Z = \dfrac{{P \times m}}{{\rho \times RT}}$
After, putting the value in the above formula, we get the value of Z, as below:
$ Z = \dfrac{{1 \times {{10}^5} \times 18 \times {{10}^{ - 3}}}}{{0.6 \times 8.314 \times 373}} \\
  Z = 0.967 \\
$
Hence the correct answer is (b) i.e. 0.967

Note:
Compressibility factor is also known as gas deviation factor or the compression factor. The other definition of compression factor is the ratio of the molar volume of a gas to the molar volume of an ideal gas at same temperature pressure condition.
$Z = \dfrac{{{V_{real}}}}{{{V_{ideal}}}}$
If the value of $Z < 1$, i.e. there is negative deviation that means there is more compressibility than the expected from ideal behaviour.
If the value of $Z > 1$, i.e. there is positive deviation that means there is lesser compressibility than the expected from ideal behaviour.
For H and He, Z is always greater than one.