To solve this problem, we can use the combined gas law, which states:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where: P1 and P2 are the initial and final pressures of the gas, V1 and V2 are the initial and final volumes of the gas, T1 and T2 are the initial and final temperatures of the gas.
Let's plug in the given values into the equation:
P1 = 700.0 mmHg V1 = 600.0 mL T1 = 200°C = 473.15 K (we need to convert from Celsius to Kelvin)
P2 = 500.0 mmHg V2 = ? T2 = -20.0°C = 253.15 K (we need to convert from Celsius to Kelvin)
Now we can solve for V2:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
(700.0 mmHg * 600.0 mL) / (473.15 K) = (500.0 mmHg * V2) / (253.15 K)
Simplifying the equation:
(700.0 mmHg * 600.0 mL * 253.15 K) = (500.0 mmHg * V2 * 473.15 K)
Dividing both sides by (500.0 mmHg * 473.15 K):
V2 = (700.0 mmHg * 600.0 mL * 253.15 K) / (500.0 mmHg * 473.15 K)
Canceling out units and performing the calculation:
V2 = (700.0 * 600.0 * 253.15) / (500.0 * 473.15) ≈ 570.38 mL
Therefore, the volume of the gas at -20.0°C and 500.0 mmHg is approximately 570.38 mL.