Question

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Answer 1

Answer:

SrCO3(s) → SrO(s) + CO2

as SrCO3 and SrO is in solid state,

Kp = PCo2

maximum pressure of CO2 = 1.6 atm

now,

P1V1 = P2V2 (as μ , R and T is constant)

putting the values of P and V given in question

0.4 × 20 = 1.6 × V2

V2 = 5L

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.