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Step-by-Step Solution
Step 1: Identify the Chemical Equilibrium
The given reaction is:
SrCO3(s) ⇌ SrO(s) + CO2(g).
Its equilibrium constant in terms of pressure is
Kp = 1.6 atm, which corresponds to the equilibrium partial pressure of CO2.
Step 2: Note the Initial Conditions
• Initial pressure of CO2, P1 = 0.4 atm
• Initial volume, V1 = 20 L
• Temperature, T = 400 K
• The container is sealed, so there is no exchange of gas with the surroundings. The number of moles of CO2 remains constant throughout the process.
Step 3: Relate Pressure, Volume, and Moles of CO₂
Using the ideal gas equation at the initial state:
$n = \frac{P_1 \, V_1}{R \, T}$.
Here, n is the number of moles of CO2 in the container.
Step 4: Maximum Partial Pressure of CO₂ at Equilibrium
The maximum attainable pressure of CO2 in the container is given by the equilibrium constant
$K_p = 1.6 \,\text{atm}.$
Thus, when the system is at equilibrium with SrO and SrCO3, the pressure of CO2 cannot exceed 1.6 atm.
Step 5: Find the Final Volume for P₂ = 1.6 atm
At equilibrium pressure P2 = 1.6 atm, the number of moles is still n. So, the ideal gas equation gives:
$V_2 = \frac{n \, R \, T}{P_2}.$
Step 6: Substitute the Value of n
From Step 3, $n = \frac{P_1 \, V_1}{R \, T}.$
Substituting this into the expression for V2, we get:
$V_2 = \left(\frac{P_1 \, V_1}{R \, T}\right) \times \frac{R \, T}{P_2}
= \frac{P_1 \, V_1}{P_2}.$
Plugging in the values:
$V_2 = \frac{0.4 \times 20}{1.6} = 5 \,\text{L}.$
Final Answer
The maximum volume of the container, when the pressure of CO2 attains its maximum value of 1.6 atm, is
5 L.
