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The compressibility of a gas is less than unity at STP. Therefore, a. Vm > 22.4 L b. Vm < 22.4 L c. Vm = 22.4 L Vm = 44.8 L

Shane Macguire , 10 Years ago
Grade upto college level
anser 3 Answers
Aditi Chauhan

Last Activity: 10 Years ago

Sol. Compressibility factor (Z) = (V )/V_id < 1 (given) ⇒ V < 22.4 L ∵ Vid (1 mol) = 22.4 L at STP

satyendra kumar

Last Activity: 7 Years ago

z=molar volume of real gas/molar volume of ideal gas *molar volume of ideal gas at STP is 22.4L*for z less than 1,molr volume of real gass should be less than 22.4L

Hitanshu Sekhar

Last Activity: 7 Years ago

Given that, Z ( compressibility factor ) of a gas
We know that, Z = pV/nRT
For an ideal gas Z=1
At high pressure value of Z increases because repulsion b/w molecules gives a noticable effect making molar volume Vm of the real gas greater than the molar volume of the ideal gas (Vm (ideal gas ) = RT/p ) causing Z to exceed 1.
 
When pressures are lower, attractive forces dominate making Z
So, if we increase the molar volume Vm of a gas, the pressure decreases, as a result attractive forces dominate making Z
 
So, the answer is Vm>22.4L. 

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