The volume of a gas collected when the temperature is 20.0°C and the pressure is 720 mm Hg
measures 0.0228 L. What is the calculated volume of the gas at 28.0°C and 780 mm Hg?

Explanation:

We can use the Combined Gas Laws to solve this problem

Data

p₁ = 720 Torr;  V₁ = 0.0228 L; T₁ = 20.0 °C  

p₂ = 780 Torr; V₂ = ?;              T₂ = 28.0 °C

Calculations

(a) Convert the temperatures to kelvins

T₁ = (20.0 + 273.15) K = 293.15 K

T₂ = (28.0 + 273.15) K = 301.15 K

(b) Calculate the new volume

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%