The pressure and temparature of an ideal gas in a closed vessel are 720 kPa and `40^(@)`C respectively. If `(1)/(4)`the of the gas is released from the vessel and the temperature of the remaining gas is raised to `353^(@)C`, the final pressure of the gas is.

Options

A. 1440 kPa
B. 1080kPa
C. 720 kPa
D. 540kPa

Explanation

To solve the problem, we will use the ideal gas law and the concept of pressure, volume, and temperature relationships.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial Pressure, \( P_1 = 720 \, \text{kPa} \)
- Initial Temperature, \( T_1 = 40^\circ C = 40 + 273 = 313 \, \text{K} \)

2. **Calculate the Initial Amount of Gas:**
- Let the initial amount of gas be \( n \) moles. The volume of the gas is constant since it is in a closed vessel.

3. **Release of Gas:**
- When \( \frac{1}{4} \) of the gas is released, the remaining amount of gas is:
\[ n' = n - \frac{1}{4}n = \frac{3}{4}n \]

4. **New Temperature:**
- The temperature of the remaining gas is raised to \( T_2 = 353^\circ C = 353 + 273 = 626 \, \text{K} \)

5. **Using the Ideal Gas Law:**
- The ideal gas law states that \( PV = nRT \). Since the volume \( V \) and the gas constant \( R \) remain constant, we can set up the relationship:
\[ \frac{P_1 \cdot n}{T_1} = \frac{P_2 \cdot n'}{T_2} \]
- Substituting \( n' = \frac{3}{4}n \):
\[ \frac{P_1 \cdot n}{T_1} = \frac{P_2 \cdot \frac{3}{4}n}{T_2} \]

6. **Simplifying the Equation:**
- We can cancel \( n \) from both sides:
\[ \frac{P_1}{T_1} = \frac{P_2 \cdot \frac{3}{4}}{T_2} \]
- Rearranging gives:
\[ P_2 = \frac{P_1 \cdot T_2}{T_1} \cdot \frac{4}{3} \]

7. **Substituting Values:**
- Substitute \( P_1 = 720 \, \text{kPa} \), \( T_1 = 313 \, \text{K} \), and \( T_2 = 626 \, \text{K} \):
\[ P_2 = \frac{720 \cdot 626}{313} \cdot \frac{4}{3} \]

8. **Calculating \( P_2 \):**
- First calculate \( \frac{720 \cdot 626}{313} \):
\[ = \frac{450720}{313} \approx 1440 \, \text{kPa} \]
- Now multiply by \( \frac{4}{3} \):
\[ P_2 = 1440 \cdot \frac{4}{3} = 1920 \, \text{kPa} \]

### Final Answer:
The final pressure of the gas is approximately \( 1920 \, \text{kPa} \).

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