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A gas obeys the equation of state P(V-b) = RT (The parameter b is a constant). The slope for an isochore will be

$(a)\;negative\qquad(b)\;zero\qquad(c)\;\large\frac{R}{(V-b)}\qquad(d)\;\large\frac{R}{P}$

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A)
$P(V-b) = RT$
$P = \large\frac{RT}{(V-b)}$
$P = (\large\frac{R}{(V-b)})T+0$
Y = mx+c(Y = P and X = T for isochore)
$Slope = (\large\frac{R}{(V-b)})$
Hence answer is (c)
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