A sample of air has a volume of 140.0 mL at 67 degrees Celsius. At what temperature would its volume be 50.0 mL at constant pressure?
(V1/T1_) = (V2/T2)
Don't forget T must be in Kelvin.
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a gas sample. The equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the pressures of the gas (since pressure is constant, it cancels out),
V1 and V2 are the volumes of the gas, and
T1 and T2 are the temperatures of the gas in Kelvin.
First, we need to convert the initial and final volumes to liters, as the equation uses SI units.
140.0 mL = 0.140 L
50.0 mL = 0.050 L
We also need to convert the initial temperature from Celsius to Kelvin. The Kelvin scale is 273.15 degrees higher than the Celsius scale, so we add 273.15 to the given temperature.
67 degrees Celsius + 273.15 = 340.15 K
Now, we can plug in the values into the equation and solve for T2:
(1 * 0.140) / (340.15) = (1 * 0.050) / (T2)
0.140 / 340.15 = 0.050 / T2
0.000411244 = 0.050 / T2
To isolate T2, we can cross-multiply:
0.000411244 * T2 = 0.050
T2 = 0.050 / 0.000411244
T2 ≈ 121.61 K
Finally, we round the answer to the appropriate number of significant figures. In this case, since the given volumes have three significant figures, we round the temperature to three significant figures as well.
Therefore, the temperature at which the volume would be 50.0 mL at constant pressure is approximately 121.6 K.