Question

For compressibility factor, Z, which of the following is /are correct?

• A. For most of the real gases Z decreases with P at the lower pressure
• B. For most of the real gases Z increases with P at the higher pressure
• C. For $H_2$ (g) and $He\left(g\right)$ Z increases with P at all the pressure & at room temperature
• D. All of the above statements are correct

Answer 1

The correct option is D All of the above statements are correct

(a) For 1 mole van der Waals equation is
$(P+\frac{A}{V_m^2})(V_m-b)=RT$
At low pressure
$Z=1-\frac{a}{V_{m}RT}(V_m\propto \frac{1}{P})$
So Z decreases with P
(b) At higher pressure
$Z=1+\frac{Pb}{RT}$
So Z increase with P
(c) For $H_2\left(g\right)$ & He(g) value of 'a' is extremely small so they are difficult to liquefy. Thus
$Z=1+\frac{Pb}{RT}$
Here Z increases with P at all the pressure.

Theory:

Compressibility Factor $\left(Z\right)$

Ratio of the observed volume of the gas to the calculated volume at the same n, T and P

$Z=\left(\frac{V_{real}}{V_{initial}}\right)$

Also, $(PV{)}_{ideal}=nRT$

$Z=\frac{P{V}_{real}}{nRT}=\frac{P{V}_{m,real}}{RT}$

$V_m=\frac{V}{n}$

Here Molar Volume $V_m$

Suppose that $Z=\frac{V_{m,real}}{V_{m,ideal}}$

$Z=1$ then $V_{real}=V_{ideal}$ then $PV=nRT$ Ideal gas

$Z\ne 1$ then $V_{real}\ne V_{ideal}themPV\ne nRT$ Real gas

$Z$ and deviation from Ideal Behaviour

$Z<1$ shows negative deviation and $V_m<22.4L$ at STP

$Z>1$ show positive deviation and $V_m>22.4L$ at STP

At origin pressure is negligible. Therefore $V_{real}=V_{ideal}$

Compression is applied then at bottom most point $V_{real}<V_{ideal}$, attraction between the gas molecules is maximum.

Further compression beyond the dip point will lead to an increase in repulsive forces among the molecules which is more than an increase in the attractive force.

Net attraction starts to decrease.

When $Z=1$ the attractive forces = repulsive forces

When $Z>1$ the attractive forces < repulsive forces (Gas difficult to compress)

When $Z<1$ the attractive forces > repulsive forces (Gas easy to compress)

Plot of $ZvsP$ consider different gases with the same temperature so they have the same kinetic energy.

$K.E=\frac{1}{2}m{v}^2$,

Kinetic energy has been kept the same. If mass is higher, the velocity is lesser. The heaviest gas is going to have the lowest value of $v$. The heaviest gases have the largest dip.

Therefore the heaviest gas will be the slowest.

Exceptional behaviour of $HandHe$ is seen. This is because $Z>1$ repulsive forces are predominant. As $H_2$ and $He$ are small in size, their nucleus has a good hold on their electron clouds. So polarisability of them is very weak. Hence Vanderwaal forces are negligible here. Therefore gases don't show a curve as pressure is increased.

Real gas behaves like an ideal gas at low pressure and high temperature