Six grams of oxygen gas at a temperature of 30 °C have a volume of 0,00142 m³. Calculate: a) The pressure in the gas if the gas constant for oxygen is 280 J/kg K. b) The final volume of the gas if the pressure is tripled and the temperature increases to 130 °C.

Mass, $m = 6 \text{ g} = 0.006 \text{ kg}$
Temperature, $T_1 = 30^\circ\text{C}$
Volume, $V_1 = 0.00142 \text{ m}^3$
Gas constant for oxygen, $R = 280 \text{ J/kg K}$
Convert the temperature to Kelvin:

$$T_1 = 30^\circ\text{C} + 273.15 = 303.15 \text{ K}$$

We will use the Ideal Gas Law: $P_1V_1 = mRT_1$.

$$P_1 = \frac{(0.006 \text{ kg}) \times (280 \text{ J/kg K}) \times (303.15 \text{ K})}{0.00142 \text{ m}^3}$$ $$P_1 = \frac{509.292}{0.00142} \text{ Pa}$$ $$P_1 \approx 358656.338 \text{ Pa}$$

Convert to kPa and round to one decimal place:

$$P_1 \approx 358.7 \text{ kPa}$$ $$P_1 \approx \boxed{\text{358.7 kPa}}$$

Initial pressure, $P_1 = 358.656 \text{ kPa}$ (using the unrounded value from part a)
Initial volume, $V_1 = 0.00142 \text{ m}^3$
Initial temperature, $T_1 = 303.15 \text{ K}$
Final pressure, $P_2 = 3 \times P_1 = 3 \times 358.656 \text{ kPa} = 1075.968 \text{ kPa}$
Final temperature, $T_2 = 130^\circ\text{C}$
Convert the final temperature to Kelvin:

$$T_2 = 130^\circ\text{C} + 273.15 = 403.15 \text{ K}$$

We will use the Combined Gas Law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.

$$V_2 = \frac{(358.656 \text{ kPa}) \times (0.00142 \text{ m}^3) \times (403.15 \text{ K})}{(1075.968 \text{ kPa}) \times (303.15 \text{ K})}$$ $$V_2 = \frac{204.887}{326.066} \text{ m}^3$$ $$V_2 \approx 0.0005999 \text{ m}^3$$

Rounding to five decimal places:

$$V_2 \approx \boxed{\text{0.00060 m}^3}$$