Question

a cylinder of oxygen gas contains 91.3 g of o2. if the volume of the cylinder is 8.58 l, what is the pressure of the o2 in atm if the gas temperature is 21 oc?

Answer 1

To calculate the pressure of oxygen gas in a cylinder, you first convert the given mass to moles, adjust the temperature to Kelvin, and then apply the ideal gas law (PV=nRT). Using the constants and given values, the pressure is calculated to be approximately 0.912 atm.

Step-by-step explanation:

To find the pressure of the O2 in the cylinder, we would use the ideal gas law PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the mass of oxygen to moles by dividing by the molar mass of O2, which is 32.00 g/mol (since O2 consists of 2 oxygen atoms, each with an approximate atomic mass of 16.00 g/mol).

Number of moles (n) = mass (m) / molar mass (M) = 91.3 g / 32.00 g/mol = 2.853 moles of O2.

Next, convert the temperature to Kelvin: T(K) = 21 °C + 273.15 = 294.15 K.

Finally, use the ideal gas constant R = 0.0821 L·atm/mol·K and solve for P:

PV = nRT => P = (nRT) / V = (2.853 mol × 0.0821 L·atm/mol·K × 294.15 K) / 8.58 L

P = 0.912 atm (Answer rounded to three significant digits)

Answer 2

p = 8 atm

Step-by-step explanation:

m (O2) = 91.3 g

M (O2) = 32 g/mol

V = 8.58 L

T = 21 °C = 294 K

R = 0.082(l x at)/(mol x K)

m = n x M

n(O2) = 91.3 g /32 g/mol = 2.85 mol

pV = nRT --> p = nRT/V = 2.85 x 0.082 x 294 / 8.58 =

= 8 atm