A balloon is filled at $27^{\circ} \mathrm{C}$ and 1 atmospheric pressure by volume $500 \mathrm{~m}^3$ helium gas. At $-3^0 \mathrm{C}$ and 0.5 atmospheric pressure, the volume of helium gas will be

The volume is determined using the combined gas law, which for a fixed amount of gas gives $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

Convert temperatures to the absolute scale: $T_1 = 27^\circ\text{C} + 273 = 300\text{ K}$ and $T_2 = -3^\circ\text{C} + 273 = 270\text{ K}$.

Substitute the known values $P_1 = 1\text{ atm}$, $V_1 = 500\text{ m}^3$, and $P_2 = 0.5\text{ atm}$, then solve for $V_2$:

$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1 \times 500 \times 270}{0.5 \times 300} = \frac{135000}{150} = 900\text{ m}^3$.

The helium occupies 900 m³ under the new conditions.