Two solid `A "and" B` are present in two different container having same volume and same temperature following equilibrium are established:
In container `(1)A(s)hArrD(g)+C(g)P_(T)=40 "atm at equilibrium"`
In container `(2)B(s)hArrE(g)+F(g)P_(T)=60 "atm at equilibrium"`
If excess of `A "and" B` are added to a third container having double the volume and at same temperature then, the total pressure of this container at equilibrium is:

Options

A. `50` atm
B. `100` atm`
C. `200` atm
D. `70` atm

Answer

To solve the problem step by step, we will analyze the equilibrium conditions in both containers and then apply the principles of gas laws to find the total pressure in the third container.

### Step 1: Understand the Equilibrium Conditions
In the first container, the equilibrium established is:
\[ A(s) \rightleftharpoons D(g) + C(g) \]
At equilibrium, the total pressure \( P_1 = 40 \, \text{atm} \).

In the second container, the equilibrium is:
\[ B(s) \rightleftharpoons E(g) + F(g) \]
At equilibrium, the total pressure \( P_2 = 60 \, \text{atm} \).

### Step 2: Determine the Contribution of Each Gas to the Total Pressure
From the first container:
- Let the partial pressure of \( D \) be \( P_D \) and the partial pressure of \( C \) be \( P_C \).
- Therefore, \( P_D + P_C = 40 \, \text{atm} \).

From the second container:
- Let the partial pressure of \( E \) be \( P_E \) and the partial pressure of \( F \) be \( P_F \).
- Therefore, \( P_E + P_F = 60 \, \text{atm} \).

### Step 3: Combine the Two Containers into a Third Container
Now, we add excess amounts of \( A \) and \( B \) to a third container with double the volume (i.e., \( 2V \)) at the same temperature.

### Step 4: Calculate the Total Pressure in the Third Container
Using the ideal gas law, we know that:
\[ P_1 V + P_2 V = P_f (2V) \]
Where \( P_f \) is the final pressure in the third container.

Substituting the values:
\[ 40 \, \text{atm} \cdot V + 60 \, \text{atm} \cdot V = P_f \cdot (2V) \]

This simplifies to:
\[ (40 + 60) V = P_f \cdot (2V) \]
\[ 100V = P_f \cdot 2V \]

### Step 5: Solve for \( P_f \)
Dividing both sides by \( 2V \):
\[ P_f = \frac{100V}{2V} = 50 \, \text{atm} \]

### Final Answer
The total pressure of the third container at equilibrium is:
\[ \boxed{50 \, \text{atm}} \]