Question

A 34 cubic dm cylinder contains 212 g of oxygen gas at 21° C. What mass of oxygen must be released to reduce the pressure in cylinder to 1.24 bar?

Answer 1

Given -

P = 1.24 bar
V = 34 cubic dm
T = 294 K
R = 0.083 cubic dm bar / K mol

Using ideal gas equation,
PV = nRT
n = PV/RT
n = 1.24 x 34/ 0.083 x 294
n = 1.727 mol

Mass of oxygen present initially = 212 g
Mass of oxygen present now = 1.727 x 32 = 55.26 g

So mass of oxygen released = 212 - 55.26 = 156.74 g

Hope This Helps You!

Answer 2

¤¤ Hello!!
¤¤ Here we go!!

¤¤ Using PV = nRT
¤¤ Here!

¤¤ P = pressure
¤¤ V = volume
¤¤ n = number of moles
¤¤ R = gas constant (0.083)
¤¤ T = temperature

¤¤ n = PV/RT
¤¤ n = 1.24 x 34/ 0.083 x 294
¤¤ therefore, n = 1.727 mol

¤¤ Initial mass of oxygen = 212 g
¤¤ Final mass of oxygen = mol x mass of O2
= 1.727 x 32 = 55.26 g

¤¤ Oxygen Released = 212 - 55.26 = 156.74 g

¤¤Hope it helps!! ¤¤