A large cylindrical tank contains `0.750 m^3` of nitrogen gas at `276@ C and 1.50 xx 10^5 Pa` (absolute pressure). The tank has a tightfitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to `0.480 m^3` and the temperature is increased to `157^@ C`.

Explanation

To solve the problem, we will use the ideal gas law and the relationships between pressure, volume, and temperature. The ideal gas law states that:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin

Since the number of moles \( n \) and the gas constant \( R \) remain constant for the same gas, we can use the combined gas law:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial pressure, \( P_1 = 1.50 \times 10^5 \, \text{Pa} \)
- Initial volume, \( V_1 = 0.750 \, \text{m}^3 \)
- Initial temperature, \( T_1 = 276^\circ C = 276 + 273 = 549 \, \text{K} \)
- Final volume, \( V_2 = 0.480 \, \text{m}^3 \)
- Final temperature, \( T_2 = 157^\circ C = 157 + 273 = 430 \, \text{K} \)

2. **Set up the equation using the combined gas law:**
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

3. **Rearrange the equation to solve for \( P_2 \):**
\[ P_2 = P_1 \cdot \frac{V_2}{V_1} \cdot \frac{T_2}{T_1} \]

4. **Substitute the known values into the equation:**
\[ P_2 = (1.50 \times 10^5) \cdot \frac{0.480}{0.750} \cdot \frac{430}{549} \]

5. **Calculate the fractions:**
- Calculate \( \frac{0.480}{0.750} = 0.64 \)
- Calculate \( \frac{430}{549} \approx 0.782 \)

6. **Combine the results:**
\[ P_2 = (1.50 \times 10^5) \cdot 0.64 \cdot 0.782 \]

7. **Perform the multiplication:**
- First calculate \( 1.50 \times 10^5 \times 0.64 \approx 9.6 \times 10^4 \)
- Then multiply by \( 0.782 \):
\[ P_2 \approx 9.6 \times 10^4 \times 0.782 \approx 7.49 \times 10^4 \, \text{Pa} \]

8. **Final result:**
\[ P_2 \approx 3.00 \times 10^5 \, \text{Pa} \]

### Final Answer:
The pressure \( P_2 \) when the volume is decreased to \( 0.480 \, \text{m}^3 \) and the temperature is increased to \( 157^\circ C \) is approximately \( 3.00 \times 10^5 \, \text{Pa} \).