An ideal gas initially at pressure 1 bar is being compressed from `30m^(3)` to `10m^(3)` volume and its temperature decreases from 320 K to 280 K then find final pressure of gas.

Options

A. 2.625 bar
B. 3.4 bar
C. 1.325 bar
D. 4.5 bar

Explanation

To solve the problem of finding the final pressure of an ideal gas undergoing compression, we can use the ideal gas law, which states that the ratio of pressure (P), volume (V), and temperature (T) remains constant for a given amount of gas.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial pressure, \( P_1 = 1 \, \text{bar} \)
- Initial volume, \( V_1 = 30 \, \text{m}^3 \)
- Initial temperature, \( T_1 = 320 \, \text{K} \)
- Final volume, \( V_2 = 10 \, \text{m}^3 \)
- Final temperature, \( T_2 = 280 \, \text{K} \)

2. **Use the Ideal Gas Law:**
According to the ideal gas law, we can write:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearranging this equation to solve for \( P_2 \):
\[ P_2 = \frac{P_1 V_1 T_2}{V_2 T_1} \]

3. **Substitute the Known Values:**
Plugging in the values we have:
\[ P_2 = \frac{(1 \, \text{bar}) \times (30 \, \text{m}^3) \times (280 \, \text{K})}{(10 \, \text{m}^3) \times (320 \, \text{K})} \]

4. **Calculate the Numerator and Denominator:**
- Numerator:
\[ 1 \times 30 \times 280 = 8400 \, \text{bar} \cdot \text{m}^3 \cdot \text{K} \]
- Denominator:
\[ 10 \times 320 = 3200 \, \text{m}^3 \cdot \text{K} \]

5. **Final Calculation:**
Now, substituting back into the equation:
\[ P_2 = \frac{8400}{3200} = 2.625 \, \text{bar} \]

### Final Answer:
The final pressure of the gas is \( P_2 = 2.625 \, \text{bar} \).

---