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Question

If PPc=x (reduced pressure), VVc=y (reduced volume) and TTc=z (reduced temperature), then,

A
(x3y2)(3y1)=8z
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B
(x+3y2)(y1)=z
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C
(x3y2)(3y+1)=8z
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D
(x+3y2)(3y1)=8z
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Solution

The correct option is D (x+3y2)(3y1)=8z
As we know,
Critical pressure, Pc=a27b2
Critical volume, Vc=3b
Critical temperature, Tc=8a27Rb
For one mole of a real gas
(P+aV2)(Vb)=RT ...(i)
PPc=x (given)
P=xPc=ax27b2
VVc=y (given)
V=yVc=3by
TTc=z (given)
T=zTc=8az27Rb
Putting values of P,V and T into Equation (i), we get
(ax27b2+a(3by)2)(3byb)=R×8az27Rb
(ax27b2+a9b2y2)(3byb)=R×8az27Rb
a27b2(x+3y2)×b(3y1)=R×8az27Rb
(x+3y2)(3y1)=8z

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