A steel cylinder contains 128 grams of oxygen (O2) at 25*C and a pressure of 1000 torr. at STP 1 mol of a gas occupies 22.4 L. What is the volume in liters of the oxygen gas in the cylinder?

I would use PV = nRT

74.38L

To find the volume of oxygen gas in the cylinder, we can make use of the ideal gas law, which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature

In this case, we are given:
P = 1000 torr
T = 25 °C (which needs to be converted to Kelvin)

First, let's convert the temperature to Kelvin by adding 273.15:
T(K) = 25 + 273.15 = 298.15 K

Next, we need to find the number of moles of gas using the given mass of oxygen (O2). To do this, we need the molar mass of oxygen, which is 32 g/mol.

Number of moles (n) = mass / molar mass
n = 128 g / 32 g/mol
n = 4 mol

Now, we know the values of P, n, and T, so we can rearrange the ideal gas law to solve for V:
V = (nRT) / P

Substituting the given values into the equation:
V = (4 mol * 0.0821 L·atm/mol·K * 298.15 K) / 1000 torr

Now, we can convert torr to atm:
1 atm = 760 torr
V = (4 mol * 0.0821 L·atm/mol·K * 298.15 K) / (1000 torr / (760 torr/atm))

V = (4 * 0.0821 * 298.15) / (1000/760)
V = 0.993 L

Therefore, the volume of oxygen gas in the cylinder is approximately 0.993 liters.