To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample.
The combined gas law equation is as follows:
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
where: P₁ and P₂ are the initial and final pressures, respectively, V₁ and V₂ are the initial and final volumes, respectively, T₁ and T₂ are the initial and final temperatures in Kelvin, respectively.
Given: P₁ = 700.0 mmHg V₁ = 600.0 mL T₁ = 27°C = 27 + 273.15 = 300.15 K P₂ = 500.0 mmHg T₂ = -20.0°C = -20 + 273.15 = 253.15 K
Let's plug in the values into the combined gas law equation and solve for V₂:
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
(700.0 mmHg × 600.0 mL) / 300.15 K = (500.0 mmHg × V₂) / 253.15 K
Simplifying the equation:
(700.0 mmHg × 600.0 mL × 253.15 K) = (500.0 mmHg × V₂ × 300.15 K)
Dividing both sides by (500.0 mmHg × 300.15 K):
(700.0 mmHg × 600.0 mL × 253.15 K) / (500.0 mmHg × 300.15 K) = V₂
Calculating the value:
V₂ ≈ 503.0 mL
Therefore, the volume of the gas at -20.0°C and 500.0 mmHg is approximately 503.0 mL.