A sample of oxygen, O2, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?
48
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of temperature, pressure, and volume.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (452 torr)
V1 = initial volume (32.2 mL)
T1 = initial temperature (30 °C + 273.15) in Kelvin
P2 = final pressure (452 torr)
V2 = volume we need to find
T2 = final temperature (-70 °C + 273.15) in Kelvin
Now, let's plug in the given values into the equation:
(452 torr * 32.2 mL) / (30 °C + 273.15) = (452 torr * V2) / (-70 °C + 273.15)
Simplifying the equation, we have:
(452 torr * 32.2 mL * (-70 °C + 273.15)) / (30 °C + 273.15) = 452 torr * V2
Now, let's solve for V2 by isolating it:
V2 = (452 torr * 32.2 mL * (-70 °C + 273.15)) / (30 °C + 273.15)
Calculating this expression will give us the volume of oxygen at -70 °C and the same pressure.
To solve this problem, we can use the combined gas law formula which relates the initial and final conditions of temperature, pressure, and volume. The formula is given 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, and
T1 and T2 are the initial and final temperatures respectively.
Let's use this formula to find the final volume:
Given:
P1 = 452 torr
V1 = 32.2 mL
T1 = 30 °C + 273.15 (converting to Kelvin) = 303.15 K
T2 = -70 °C + 273.15 (converting to Kelvin) = 203.15 K
Plug in the given values into the formula:
(452 torr * 32.2 mL) / (303.15 K) = (P2 * V2) / (203.15 K)
Now, isolate V2 to solve for the final volume:
V2 = [(452 torr * 32.2 mL * 203.15 K) / (303.15 K)] / P2
Since the pressure is the same, P2 = 452 torr. Substituting this value:
V2 = [(452 torr * 32.2 mL * 203.15 K) / (303.15 K)] / 452 torr
Simplifying the equation:
V2 ≈ (32.2 mL * 203.15 K) / (303.15 K)
Calculating, we get:
V2 ≈ 21.6 mL
Therefore, the oxygen sample will occupy approximately 21.6 mL at -70 °C and the same pressure.