what is the volume (in mL) when 280 mL of an ideal gas at 87 degree C and 650 torr preasure is cooled at 27 degree C at 760 torr?
To solve this question, we can use the combined gas law and assume the gas behaves ideally. The combined gas law equation is as follows:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
where:
P₁ and P₂ are the initial and final pressures,
V₁ and V₂ are the initial and final volumes, and
T₁ and T₂ are the initial and final temperatures.
Let's plug in the given values into the equation and solve for V₂.
P₁ = 650 torr
V₁ = 280 mL
T₁ = 87 °C + 273.15 = 360.15 K
P₂ = 760 torr
T₂ = 27 °C + 273.15 = 300.15 K
Plugging the values into the equation:
(650 torr * 280 mL) / (360.15 K) = (760 torr * V₂) / (300.15 K)
Next, we can solve for V₂:
V₂ = (650 torr * 280 mL * 300.15 K) / (760 torr * 360.15 K)
V₂ ≈ 261.28 mL
Therefore, the volume of the gas when it is cooled to 27 °C at 760 torr pressure is approximately 261.28 mL.