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 step by step, we will use the ideal gas law and the relationships between pressure, volume, temperature, and the number of moles of gas.

### Step 1: Understand the 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 \( n_1 = n \).

### Step 2: Determine the final conditions after releasing half the gas
- After releasing half of the gas, the number of moles becomes \( n_2 = \frac{n}{2} \).

### Step 3: Raise the temperature of the remaining gas
- The final temperature \( T_2 = 27^\circ C + 50^\circ C = 77^\circ C = 77 + 273 = 350 \) K.

### Step 4: Apply the ideal gas law
The ideal gas law states that:
\[ PV = nRT \]
Since the volume \( V \) is constant, we can use the relationship:
\[ \frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2} \]
From this, we can rearrange to find \( P_2 \):
\[ P_2 = P_1 \cdot \frac{n_2}{n_1} \cdot \frac{T_2}{T_1} \]

### Step 5: Substitute the known values
Substituting the values we have:
- \( P_1 = 20 \) atm
- \( n_2 = \frac{n}{2} \)
- \( n_1 = n \)
- \( T_2 = 350 \) K
- \( T_1 = 300 \) K

Now substituting into the equation:
\[ P_2 = 20 \cdot \frac{\frac{n}{2}}{n} \cdot \frac{350}{300} \]
This simplifies to:
\[ P_2 = 20 \cdot \frac{1}{2} \cdot \frac{350}{300} \]
\[ P_2 = 10 \cdot \frac{350}{300} \]

### Step 6: Calculate \( P_2 \)
Calculating \( \frac{350}{300} \):
\[ \frac{350}{300} = \frac{35}{30} = \frac{7}{6} \]
Now substituting back:
\[ P_2 = 10 \cdot \frac{7}{6} = \frac{70}{6} \approx 11.67 \text{ atm} \]

### Final Answer
The final pressure of the gas after releasing half of it and raising the temperature is approximately \( 11.67 \) atm.