Study the graph between partial pressure and mole fraction of some gases and arrange the gases P, Q, R, and S dissolved in H2O, in the decreasing order of their KH values.
The given problem involves the interpretation of a graph showing the relationship between partial pressure and mole fraction for different gases (P, Q, R, and S) dissolved in water (H\(_2\)O).
The graph represents Henry’s law, which states that the solubility of a gas in a liquid is directly proportional to its partial pressure, and the proportionality constant is the Henry’s law constant \( K_H \).
From the graph: - The steeper the slope, the lower the Henry’s law constant \( K_H \), indicating a greater solubility.
- Conversely, the flatter the slope, the higher the \( K_H \), indicating a lower solubility.
By analyzing the graph, we observe that: - Gas S has the flattest slope, which corresponds to the highest \( K_H \) value (least soluble).
- Gas P has a steeper slope than gas S but less steep than gases R and Q.
- Gas R has a steeper slope than gas P, indicating a lower \( K_H \) than gas P but higher than gas Q. - Gas Q has the steepest slope, which indicates the lowest \( K_H \) value (most soluble).
Thus, the gases should be arranged in decreasing order of \( K_H \) as: \[ S>P>R>Q \] Hence, the correct answer is \( (A) \).
A binary mixture of benzene and toluene under vapour-liquid equilibrium at 80 \( ^\circ {C} \) follows ideal Raoult’s law. At this condition, the saturation pressures of benzene and toluene are 101 kPa and 40 kPa, respectively. If the mole fraction of benzene in the liquid phase is 0.6, the corresponding mole fraction of benzene in the vapour phase will be ________. (Round off to two decimal places)
A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 5 rad/s. A particle of mass 1 kg moving with a velocity of 5 m/s strikes the cylinder and sticks to it as shown in figure.
The angular velocity of the system after the particle sticks to it will be: