Two gases occupy two containers `A` and `B` then gas in `A`, of volume `0.10 m^(3)`, exerts a pressure of `1.40 Mpa` and that in `B`, of volume `0.15 m^(3)`, exerts a pressure `0.7 MP a`. The two containers are joined by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

Options

A. 0.7
B. 0.98
C. 1.4
D. 2.1

Explanation

To solve the problem, we will use the concept of the ideal gas law and the principle of conservation of moles when the two gases are allowed to intermingle.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Volume of gas in container A, \( V_A = 0.10 \, m^3 \)
- Pressure of gas in container A, \( P_A = 1.40 \, MPa \)
- Volume of gas in container B, \( V_B = 0.15 \, m^3 \)
- Pressure of gas in container B, \( P_B = 0.70 \, MPa \)

2. **Use the Ideal Gas Law:**
The number of moles of gas can be calculated using the ideal gas law, which states:
\[ n = \frac{PV}{RT} \]
Since the temperature \( T \) and the gas constant \( R \) are constant, we can express the number of moles in each container as:
\[ n_A = \frac{P_A V_A}{RT} \quad \text{and} \quad n_B = \frac{P_B V_B}{RT} \]

3. **Total Number of Moles After Mixing:**
The total number of moles after the gases intermingle will be:
\[ n_{total} = n_A + n_B = \frac{P_A V_A}{RT} + \frac{P_B V_B}{RT} \]
This can be simplified to:
\[ n_{total} = \frac{1}{RT} (P_A V_A + P_B V_B) \]

4. **Final Pressure Calculation:**
The final pressure \( P_f \) after mixing can be expressed as:
\[ P_f = \frac{n_{total} RT}{V_A + V_B} \]
Substituting \( n_{total} \):
\[ P_f = \frac{(P_A V_A + P_B V_B)}{(V_A + V_B)} \]

5. **Substituting the Values:**
Now, substituting the known values:
\[ P_f = \frac{(1.40 \, MPa \times 0.10 \, m^3) + (0.70 \, MPa \times 0.15 \, m^3)}{0.10 \, m^3 + 0.15 \, m^3} \]

6. **Calculating the Numerator:**
- For container A: \( 1.40 \times 0.10 = 0.14 \, MPa \cdot m^3 \)
- For container B: \( 0.70 \times 0.15 = 0.105 \, MPa \cdot m^3 \)
- Total: \( 0.14 + 0.105 = 0.245 \, MPa \cdot m^3 \)

7. **Calculating the Denominator:**
\[ V_A + V_B = 0.10 + 0.15 = 0.25 \, m^3 \]

8. **Final Pressure Calculation:**
\[ P_f = \frac{0.245 \, MPa \cdot m^3}{0.25 \, m^3} = 0.98 \, MPa \]

### Final Answer:
The final pressure in the container after the gases intermingle is \( P_f = 0.98 \, MPa \).

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