The gas sample has a volume of 76.7 μL at 22.1 °C. What is the volume of the gas after the sample is cooled to 9.8 °C at constant pressure?
(V1/T1) = (V2/T2)
Remember T must be in kelvin.
To determine the volume of the gas after it is cooled to 9.8 °C at constant pressure, we can use the combined gas law equation.
The combined gas law equation is given by:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 and P2 are the initial and final pressures (assuming they are constant), respectively,
V1 and V2 are the initial and final volumes, respectively,
T1 and T2 are the initial and final temperatures (in Kelvin), respectively.
First, we need to convert the temperatures from Celsius to Kelvin using the formula:
Kelvin = Celsius + 273.15
So, let's convert the given temperatures:
T1 = 22.1 °C + 273.15 = 295.25 K
T2 = 9.8 °C + 273.15 = 283.95 K
Now, we can form the equation as follows:
(V1) / (T1) = (V2) / (T2)
Substituting the known values:
(76.7 μL) / (295.25 K) = (V2) / (283.95 K)
Now, we can solve for V2:
V2 = (76.7 μL) * (283.95 K) / (295.25 K)
Calculating the value on the right:
V2 ≈ 73.97 μL
Therefore, the volume of the gas after the sample is cooled to 9.8 °C at constant pressure is approximately 73.97 μL.