The gas in a vessel is subjected to a pressure of `20` atmosphere at a temperature `27^(@)C`. The pressure of the gas in the vessel after one half of the gas is released from the vessel and the temperature of the remainder is raised by `50^(@)C` is

Options

A. 8.5 atm
B. 10.8 atm
C. 11.7 atm
D. 17 atm

Explanation

To solve the problem, we will use the ideal gas law, which states that \( PV = nRT \). We will analyze the situation step by step.

### Step-by-Step Solution:

1. **Initial Conditions**:
- Initial pressure \( P_1 = 20 \) atm
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \) K
- Let the initial number of moles of gas be \( n_1 = n \).
- The volume of the vessel is \( V \).

2. **After Releasing Half the Gas**:
- When half of the gas is released, the number of moles left is:
\[ n_2 = \frac{n}{2} \]

3. **Final Temperature**:
- The temperature is raised by \( 50^\circ C \):
\[ T_2 = 27 + 50 = 77^\circ C = 77 + 273 = 350 \text{ K} \]

4. **Using the Ideal Gas Law**:
- For the initial state (before releasing gas):
\[ P_1 V = n R T_1 \tag{1} \]
- For the final state (after releasing gas and heating):
\[ P_2 V = n_2 R T_2 \tag{2} \]

5. **Substituting Known Values**:
- Substitute \( n_2 = \frac{n}{2} \) into equation (2):
\[ P_2 V = \left(\frac{n}{2}\right) R (350) \]

6. **Dividing Equation (2) by Equation (1)**:
- This gives:
\[ \frac{P_2 V}{P_1 V} = \frac{\left(\frac{n}{2}\right) R (350)}{n R (300)} \]
- Simplifying, we find:
\[ \frac{P_2}{P_1} = \frac{\frac{1}{2} \cdot 350}{300} \]

7. **Calculating the Ratio**:
- Simplifying the right side:
\[ \frac{P_2}{20} = \frac{350}{600} = \frac{7}{12} \]

8. **Finding \( P_2 \)**:
- Rearranging gives:
\[ P_2 = 20 \cdot \frac{7}{12} \]
- Calculating \( P_2 \):
\[ P_2 = \frac{140}{12} = \frac{35}{3} \approx 11.67 \text{ atm} \]

### Final Answer:
The pressure of the gas in the vessel after one half of the gas is released and the temperature of the remainder is raised by \( 50^\circ C \) is approximately \( 11.67 \) atm.

---