Answer to Question #301219 in Chemistry for Kyle

Question

A cylinder whose volume is 34.0 Liters contains 305 grams if elemental Oxygen at 22 degrees Celsius. How many grams of gas must be released to reduce the pressure to 1.15 atm if the temperature remains constant? Give complete solution.

Answer 1

P = 1.15 atm

V = 34.0 L

T = 22^∘C + 273.15 = 295.15 K

R = gas constant = 0.0821 L atm K⁻¹ mol⁻¹

Solution:

The ideal gas law can be used to determine moles of O₂

PV = nRT

Rearrange the equation to isolate n:

n = PV / RT

Plug in the known values and solve for n:

n(O₂) = (1.15 atm × 34.0 L) / (0.0821 L atm K⁻¹ mol⁻¹ × 295.15 K) = 1.6136 mol O₂

The molar mass of O₂ is 31.998 g/mol

Therefore,

Mass of O₂ = (1.6136 mol O₂) × (31.998 g O₂ / 1 mol O₂) = 51.63 g O₂

The amount of O₂ that must be released:

305 g O₂ − 51.63 g O₂ = 253.37 g O₂

Answer: 253.37 grams of gas (O₂) must be released