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The compressibillity factor for 1 mole o...

The compressibillity factor for 1 mole of a van der Waals gas at 273 K and 100 atm pressure is 0.5. Assuming that the volume of
a gas molecule is negligible, calculate the van der Waals constant a ( in units of atm `L^(2)"mol"^(-2)).`

Text Solution

Verified by Experts

The correct Answer is:
1.26
`Z=(PV)/(RT)=0.5=(100 xx V)/(0.0821 xx 273)`
`therefore V=0.112L` Now, `(P+(n^(2)alpha)/(V^(2)))(V-nb)=nRT`
`therefore (100 +(a)/(0.112 xx 0.112))(0.112-0)`
=0.0821 xx 273=22.41 `therefore((100 xx 0.112)+(a)/(0.112))=22.41`
`therefore 11.2+(a)/(0.112)=22.41`
`therefore (alpha)/(0.112)=22.41-11.2`
`therefore(alpha)/(0.112)=11.21`
`therefore a=1.26 " atm "L^(2)" mol"^(-2)`
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