Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

Hi, I'm struggling to understand these chemistry questions. Help with any or all would be greatly appreciated! Thank you!!

1. Medical oxygen contained in a 20.0L steel tank was removed at a constant temperature of 27°C. the pressure in the tank decreased from 122.0 atm to 12.6 atm. How much oxygen, in grams, was removed from the tank?

2. A balloon is filled with 600ml of helium at a a temperature of 27°C and 860 torr. As the balloon rises in the atmosphere, pressure and temperature drop. What volume will the balloon have when it reaches an altitude where the temperature is -2.6°C and the pressure is 0.600 atm?

3. If a gas in a box with a fixed volume has a pressure of 645 torr at 15°C, what temperature should the gas be heated to double the pressure?

Thank you again, I really appreciate your help!

Relevance

1.

For the removed oxygen:

Pressure, P = (122.0 - 12.6) atm = 109.4 atm

Volume, V = 20.0 L

Mass, n = ? g

Molar mass, M = 16.0×2 g/mol = 32.0 g/mol

Gas constant, R = 0.08206 L atm / (mol K)

Temperature, T = (273 + 27) K = 300 K

PV = nRT and n = m/M

Then, PV = (m/M)RT

Hence, m = PVM/(RT)

Mass, m = 109.4 × 20.0 × 32.0 / (0.08206 × 300) g = 2840 g

====

2.

Initial: P₁ = 860/760 atm, V₁ = 600 ml, T₁ = (273 + 27) K = 300 K

Final: P₂ = 0.600 atm, V₂ = ? ml, T₂ = (273 - 2.6) K = 270.4 K

Combined gas law: P₁V₁/T₁ = P₂V₂/T₂

Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)

Final volume, V₂ = (600 ml) × [(860/760) / 0.600)] × (270.4/300) = 1020 ml

====

3.

Initial: P₁ = 645 torr, T₁ = (273 + 15) K = 288 K

Final: P₂ = 645×2 torr, T₂ = ? K

Boyle’s law: P₁/T₁ = P₂/T₂

Then, T₂ = T₁ × (P₂/P₁)

Final temperature, T₁ = (288 K) × 2 = 576 K = 303°C

• Login to reply the answers