Question

If the vapour is compressed further, at what pressure, almost complete condensation of vapour will occur?
A. $0.4bar$
B. $0.5bar$
C. $0.52bar$
D. $0.6bar$

Answer 1

Hint: Condensation is the transformation of water vapour into liquid. It's the opposite of evaporation, in which liquid water turns into a mist. Condensation occurs when the air is either cooled to its dew point or becomes saturated with water vapour to the point that it can no longer carry any more water.

Complete answer:
The pressure exerted by a vapour with its condensed phases (solid or liquid) in a closed system at a given temperature is known as vapour pressure or vapour equilibrium pressure (in a system with thermodynamic equilibrium). The equilibrium vapour pressure is a well-known evaporation rate measurement. The proclivity of particles to escape from a liquid (or a solid) is thought to be associated. A product with a high vapour pressure at room temperature is referred to as a "volatile material." It's important to note that vapour pressure refers to the force that a vapour exerts on a liquid surface.
Now, coming to the question:
Vapour pressure of pure A ${P^ \circ }A = 0.4bar$
Vapour pressure of pure B ${P^ \circ }B = 0.6bar$
${X_A} = \dfrac{{Moles\,ofA}}{{Moles\,ofA + Moles\,ofB}}$
${X_B} = \dfrac{{Moles\,of\,B}}{{Moles\,ofA + Moles\,ofB}}$
${P_T}$ i.e. where the complete condensation will occur
\[
  {P_T} = {P^ \circ }A \times A + {P^ \circ }B \times B \\
  \,\,\,\,\,\,\, = \left( {0.4} \right) \times \left( {0.4} \right) + \left( {0.6} \right) \times \left( {0.6} \right) \\
  {P_T} = 0.52bar \\
 \]
Hence, complete condensation of vapour will at $0.52bar$
So, the correct option is: (c) $0.52bar$.

Note:
As the temperature of a liquid rises, so does the kinetic energy of its molecules. The number of molecules transitioning into a vapour increases as the kinetic energy of the molecules increases, raising the vapour pressure.