Question
The density of vapour of a substance (X) at 1 atm pressure and 500 K is 0.8 kgm3. The vapour effuses through a small hole at a rate of 45 times slower than oxygen under the same condition. What is the compressibility factor (Z) of the vapour?
Answer: Option C
:
C
rxrO2=√MO2Mx=⟮45⟯2=32Mx⇒Mx=50
dx=0.80kgm−3
Vm=1000800×50=62.5L
Z=PVmRT=1×62.50.0821×500=1.52
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:
C
rxrO2=√MO2Mx=⟮45⟯2=32Mx⇒Mx=50
dx=0.80kgm−3
Vm=1000800×50=62.5L
Z=PVmRT=1×62.50.0821×500=1.52
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