Question

A 20 litre container at 400 K contains CO2(g) at
pressure 0.4 atm and an excess of SO (neglect the
volume of solid SrO)The volume of the container
is now decreased by moving the movable piston
fitted in the container. The maximum volume of the
container, when pressure of Co2, attains its
maximum vol be
Given that SrCO3(s)=SrO(s)+co2(g)
Kp - 1.6atm
(1) 10
(2) 4 litre
(3) 2
(4)5 litre​

Answer 1

Answer: 5 litre

Explanation:

Answer 2

Hey there!

Given Reaction is:

$SrCO_3(s)\rightleftharpoons}SrO(s)+CO_2(g)$

Here,

$K_p = P_{CO_2}$  = 1.6 atm { ∵$SrCO_3\ and \ SrO$ are in solid state}

It is maximum pressure of CO₂ at equillibrium.

Initial volume (Vi)= 20 L

Initial Pressure(Pi) = 0.4 atm

Final Pressure(Pf) = 1.6 atm  (at equillibrium)

Let max. volume of container be Vf.

Since the process is taking place in sealed container, so n(i.e number of moles) , T (i.e Temperature) will be constant.

And as the temperature is constant so it will follow boyle's law which states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

$i.e$$P \propto \dfrac{1}{V}$

or, Pi * Vi = Pf * Vf

0.4 * 20 = 1.6 * Vf

Vf = 0.4*20/1.6

Vf = 5 L

Hence, the maximum volume of the container, when pressure of CO₂ attains its maximum value will be 5 L.