Calculate the volume occupied by 80 grams of oxygen gas at 30°C and 800 torr Hg pressure.
Answers
Answer:
The volume occupied by 80g O₂ at 30°C and 800torr is 59L
Explanation:
Given,
mass of O₂ = 80g
Pressure, P = 800torr = 800mmHg
⇒ = 800/760 = 1.053atm
Temperature, T = 30°C
⇒ = 273+30 = 303K
Gas constant, R = 0.082L.atm.K⁻¹mol⁻¹
moles of O₂, n = mass of O₂ / molar mass of O₂
⇒ = 80g / 32gmol⁻¹ = 2.5mol.
The ideal gas equation is given by,
PV = nRT
rearranging the equation,
V = nRT/P
Substituting the values of n, R, T, and P we get,
Volume of O₂ = (2.5×0.082×303) / 1.053 = 58.988L≈59L
Hence, the required answer is 59L
Answer:
The volume occupied by 80g O₂ at 30°C and 800torr is 59L
Explanation:
Given,
mass of O₂ = 80g
Pressure, P = 800torr = 800mmHg
⇒ = 800/760 = 1.053atm
Temperature, T = 30°C
⇒ = 273+30 = 303K
Gas constant, R = 0.082L.atm.K⁻¹mol⁻¹
moles of O₂, n = mass of O₂ / molar mass of O₂
⇒ = 80g / 32gmol⁻¹ = 2.5mol.
The ideal gas equation is given by,
PV = nRT
rearranging the equation,
V = nRT/P
Substituting the values of n, R, T, and P we get,
Volume of O₂ = (2.5×0.082×303) / 1.053 = 58.988L≈59L
Hence, the required answer is 59L