A 35.0-L cylinder contains 387 g O2(g) at 22.5°C. What mass of O2(g) must be released to reduce the pressure in the cylinder to 5.94 atm, assuming the temperature remains constant?
Use PV = nRT and substitute the conditions for the 5.94 atm.
p = 5.94 atm
V = 35.0 L
R = 0.08206
T = 273.15 + 22.5 = ?
Solve for n and convert n to grams.
x grams = n x molar mass O2.
You started with 387g
You want to end up with x grams
The difference is what you must vent
290.8g
To solve this problem, we need to use the Ideal Gas Law equation, which is expressed as:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal Gas Constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15 = 22.5°C + 273.15 = 295.65 K
Using the given information, we can determine the initial number of moles (n1) of O2 in the cylinder by rearranging the Ideal Gas Law equation:
n1 = PV / RT
n1 = (5.94 atm) * (35.0 L) / (0.0821 L·atm/(mol·K)) * (295.65 K)
n1 = 0.8706 mol
Next, we need to find the final number of moles (n2) of O2. Since the pressure is only changing while keeping the volume and temperature constant, we can use the ratio between the initial and final pressures to determine the final number of moles:
n2 = n1 * (P1 / P2)
n2 = 0.8706 mol * (5.94 atm / 1 atm)
n2 = 5.1691 mol
Now, to find the mass of O2 that needs to be released, we can use the molar mass of O2, which is 32.00 g/mol. The formula to calculate the mass (m) from the number of moles (n) and molar mass (M) is:
m = n * M
m = 5.1691 mol * 32.00 g/mol
m ≈ 165.42 g
Therefore, approximately 165.42 grams of O2 gas must be released to reduce the pressure in the cylinder to 5.94 atm, assuming the temperature remains constant.