Answer check: I got 1.66 x 10^3 g O2
A steel cylinder contains 98.0 moles of oxygen gas at a temperature of 38 degrees C and a pressure of 740 torr. After some of the oxygen gas has been used, the pressure is reduced to 332 toor at a temperature of 24 degrees C. What mass of oxygen gas is removed from the cylinder?
Use PV = nRT and solve for the NEW number moles from the second set of conditions and subtract from the original moles to give you moles oxygen removed. Then n = grams/molar mass and solve for grams.
To find the mass of oxygen gas removed from the cylinder, we need to calculate the difference in the number of moles of oxygen gas before and after the pressure reduction.
We can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Initially, the conditions are:
P1 = 740 torr
T1 = 38 degrees C (which needs to be converted to Kelvin)
n1 = 98.0 moles
Converting temperature to Kelvin:
T1(K) = T1(C) + 273.15
T1(K) = 38 + 273.15
T1(K) = 311.15 K
We can rearrange the ideal gas law equation to isolate n1:
n1 = (P1 * V) / (R * T1)
Now let's find n1:
n1 = (740 torr * V) / (R * 311.15 K)
Next, we have the conditions after the pressure reduction:
P2 = 332 torr
T2 = 24 degrees C (which also needs to be converted to Kelvin)
Converting temperature to Kelvin:
T2(K) = T2(C) + 273.15
T2(K) = 24 + 273.15
T2(K) = 297.15 K
We can rearrange the ideal gas law equation to isolate n2:
n2 = (P2 * V) / (R * T2)
Now, let's find n2:
n2 = (332 torr * V) / (R * 297.15 K)
The difference in moles of oxygen gas is:
Δn = n1 - n2
To convert the difference in moles to grams, we need to use the molar mass of oxygen (O2), which is 32.00 g/mol.
Mass of oxygen gas removed = Δn * molar mass of oxygen gas
Finally, plug in the values and calculate the answer.