A 33.0-L cylinder contains 481 g O2(g) at 22.5°C. What mass of O2(g) must be released to reduce the pressure in the cylinder to 6.16 atm, assuming the temperature remains constant?

Thanks! :D

Jin? Abby?

Use PV = nRT and calculate n to reduce the pressure to that indicated, then n = grams/molar mass. You know n and molar mass, solve for grams.

we're in the same class!

To solve this problem, you need to use the Ideal Gas Law equation:

PV = nRT

where:
P is the pressure,
V is the volume,
n is the number of moles,
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature in Kelvin.

First, let's convert the given temperature of 22.5°C to Kelvin by adding 273.15:
T = 22.5°C + 273.15 = 295.65 K

We are given the original pressure (P1 = 1 atm) and volume (V1 = 33.0 L) of the cylinder.

Now, let's calculate the number of moles of oxygen gas in the cylinder using the ideal gas law:
n1 = (P1 * V1) / (R * T)

Next, we need to find the new number of moles of oxygen gas when the pressure is reduced to 6.16 atm (P2 = 6.16 atm), assuming the volume and temperature remain constant:
n2 = (P2 * V1) / (R * T)

To find the mass of oxygen gas released, we need to find the difference between the initial and final number of moles:
mass of O2(g) released = (n1 - n2) * molar mass of oxygen gas

The molar mass of oxygen (O2) is 32 g/mol.

Now let's plug in the values and calculate:

Step 1:
P1 = 1 atm
V1 = 33.0 L
T = 295.65 K

n1 = (1 atm * 33.0 L) / (0.0821 L·atm/(mol·K) * 295.65 K)

Step 2:
P2 = 6.16 atm
n2 = (6.16 atm * 33.0 L) / (0.0821 L·atm/(mol·K) * 295.65 K)

Step 3:
mass of O2(g) released = (n1 - n2) * 32 g/mol

Now, let's plug in the values and calculate each step:

Step 1:
n1 = (1 atm * 33.0 L) / (0.0821 L·atm/(mol·K) * 295.65 K)
n1 ≈ 1.27 mol

Step 2:
n2 = (6.16 atm * 33.0 L) / (0.0821 L·atm/(mol·K) * 295.65 K)
n2 ≈ 6.24 mol

Step 3:
mass of O2(g) released = (n1 - n2) * 32 g/mol
mass of O2(g) released ≈ (1.27 mol - 6.24 mol) * 32 g/mol
mass of O2(g) released ≈ -177 g

Therefore, to reduce the pressure in the cylinder to 6.16 atm, approximately 177 g of O2(g) must be released, assuming the temperature remains constant.