A vessel is filled with an ideal gas at a pressure of `20 atm` and is a temperature of `27^@ C` One - half of the mass is removed from the vessel and the temperature of the remaining gas is increased to `87^@ C`. At this temperature, Find the pressure of the gas.

Explanation

To solve the problem, we will use the ideal gas law and the relationship between pressure, volume, temperature, and the number of moles of gas.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial pressure, \( P_1 = 20 \, \text{atm} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
- Final temperature, \( T_2 = 87^\circ C = 87 + 273 = 360 \, \text{K} \)
- Initial number of moles, \( n_1 \) (we'll keep it as \( n_1 \) for now)
- Final number of moles after removing half, \( n_2 = \frac{n_1}{2} \)

2. **Use the Ideal Gas Law:**
The ideal gas law states:
\[ PV = nRT \]
Since the volume \( V \) and the gas constant \( R \) remain constant, we can relate the initial and final states using:
\[ \frac{P_2}{P_1} = \frac{n_2}{n_1} \cdot \frac{T_2}{T_1} \]

3. **Substitute Values into the Equation:**
- Substitute \( n_2 = \frac{n_1}{2} \):
\[ \frac{P_2}{20} = \frac{\frac{n_1}{2}}{n_1} \cdot \frac{360}{300} \]
- Simplifying gives:
\[ \frac{P_2}{20} = \frac{1}{2} \cdot \frac{360}{300} \]

4. **Calculate the Right Side:**
- First, simplify \( \frac{360}{300} \):
\[ \frac{360}{300} = 1.2 \]
- Now substitute back:
\[ \frac{P_2}{20} = \frac{1}{2} \cdot 1.2 = 0.6 \]

5. **Solve for \( P_2 \):**
- Multiply both sides by 20:
\[ P_2 = 0.6 \cdot 20 = 12 \, \text{atm} \]

### Final Answer:
The pressure of the gas after removing half of the mass and increasing the temperature to \( 87^\circ C \) is \( P_2 = 12 \, \text{atm} \).

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