A gas at 27°C has a volume of 4 m³ and a gauge pressure of 2.202 x 10⁵ Pa. It is compressed to a volume of 0.025 m³ at 40°C. Calculate the final gauge pressure.

Initial conditions:

• Initial gauge pressure, $P_{1,gauge} = 2.202 \times 10^5 \text{ Pa}$.
• Initial volume, $V_1 = 4 \text{ m}^3$.
• Initial temperature, $T_1 = 27^\circ C$. Convert to Kelvin: $T_1 = 27 + 273 = 300 \text{ K}$.

Final conditions:

• Final volume, $V_2 = 0.025 \text{ m}^3$.
• Final temperature, $T_2 = 40^\circ C$. Convert to Kelvin: $T_2 = 40 + 273 = 313 \text{ K}$.

We need to calculate the final gauge pressure, $P_{2,gauge}$.
Assume standard atmospheric pressure, $P_{atm} = 1.013 \times 10^5 \text{ Pa}$.

Absolute pressure is the sum of gauge pressure and atmospheric pressure.

$$P_{1,abs} = P_{1,gauge} + P_{atm}$$ $$P_{1,abs} = 2.202 \times 10^5 \text{ Pa} + 1.013 \times 10^5 \text{ Pa}$$ $$P_{1,abs} = (2.202 + 1.013) \times 10^5 \text{ Pa}$$ $$P_{1,abs} = 3.215 \times 10^5 \text{ Pa}$$

The Combined Gas Law relates pressure, volume, and temperature for a fixed amount of gas:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

We need to solve for the final absolute pressure, $P_{2,abs}$. Rearrange the formula:

$$P_{2,abs} = \frac{P_{1,abs} V_1 T_2}{T_1 V_2}$$

$$P_{2,abs} = \frac{(3.215 \times 10^5 \text{ Pa}) \times (4 \text{ m}^3) \times (313 \text{ K})}{(300 \text{ K}) \times (0.025 \text{ m}^3)}$$ $$P_{2,abs} = \frac{4024.58 \times 10^5 \text{ Pa}}{7.5}$$ $$P_{2,abs} = 536.61066... \times 10^5 \text{ Pa}$$

$$P_{2,gauge} = P_{2,abs} - P_{atm}$$ $$P_{2,gauge} = (536.61066 \times 10^5 \text{ Pa}) - (1.013 \times 10^5 \text{ Pa})$$ $$P_{2,gauge} = (536.61066 - 1.013) \times 10^5 \text{ Pa}$$ $$P_{2,gauge} = 535.59766 \times 10^5 \text{ Pa}$$

Rounding to three significant figures, consistent with the input values:

$$P_{2,gauge} \approx 5.36 \times 10^7 \text{ Pa}$$

The final gauge pressure is $5.36 \times 10^7 \text{ Pa}$.