(i) Calculation of critical temperature (T_c)
a = 4.10 dm⁶ bar mol^-1
b = 0.035 dm³ mol^-1
R = 0.0821 dm³ bar mol^-1 K^-1.
Now, critical temperature, T_c = \(\frac{8a}{27Rb}\)
Substituting the values, T_c = \(\frac{8\times4.10}{27\times0.0821\times0.035}\)
(ii) Calculation of critical pressure (P_c)
P_c\(\frac{a}{27b^2}\) = \(\frac{4.10}{27\times0.0821\times0.035}\) = 123.96 bar

Given,
Initial pressure, p₁ = 1.2 bar
Initial volume, V₁ = 120 mL
Final volume, V₂ = 180 mL
Since the temperature remains constant, the final pressure (p₂) can be calculated using Boyle’s law.
According to Boyle’s law,
\( p_{1} V_{1}=p_{2} V_{2} \)
\(p_{2}=\frac{p_{1} V_{1}}{V_{2}} \)
\(=\frac{1.2 \times 120}{180} \mathrm{bar} \)
\(=0.8 \mathrm{bar} \)
Therefore, the pressure would be 0.8 bar.

When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of the upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.
In laminar flow, the velocity of molecules is not same in all the layers because every layers offers some resistance or friction to the layer immediately below it.

Graduation of velocity in the laminar flow.

Moles of O₂, ⁿO₂ = \(\cfrac{mass}{mol.wt.}=\cfrac{8}{32}=0.25mol\)
[mol.wt. of O₂ = 16\(\times\)2 = 32]
Moles of H₂, ⁿH₂ = \(\cfrac{4}{2}=2.0mol\)
[mol.wt. of H₂ = 1 \(\times\)2 = 2 ]
Total number of moles = 0.25 + 2.0 = 2.25mol
Pressure, p = \(\cfrac{nRT}{V}\)
\(\cfrac { 2.25mol\times 0.083bar \ dm^{ 3 }K^{ -1 }mol^{ -1 }\times 300K) }{ (1 \ dm^{ 3 }) } \)
p = 56.025 bar

In a closed vessel, liquid C will not boil because pressure inside keep on increasing.