The vanderwall equation of a gas is $(\begin{matrix} P + \frac{a T^{2}}{V} \end{matrix}) V^{c} = (R T + b)$ . Where $a, b, c$ and $R$ are constants. If the isotherm is represented by $P = A V^{m} - B V^{n}$ , where $A$ and $B$ depends on temperature :

(A)

m = - c, n = - 1

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(B)

m = + c, n = 1

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(C)

m = - c, n = + 1

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(D)

m = c, n = - 1

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Solution

The correct option is A (A) $m = - c, n = - 1$
The van der waal equation of a gas,
$P + \frac{a T^{2}}{V} = (R T + b) V^{- c}$

Rearranging gives
$\Rightarrow P = (R T + b) V^{- c} - a T^{2} V^{- 1} = A V^{m} - B V^{n}$

So by comparison
$\Rightarrow m = - c a n d n = - 1$

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The vanderwall equation of a gas is (P+aT2V)Vc=(RT+b). Where a,b,c and R are constants. If the isotherm is represented by P=AVm−BVn, where A and B depends on temperature :

The correct option is A (A)m=−c,n=−1The van der waal equation of a gas,P+aT2V=(RT+b)V−cRearranging gives⇒P=(RT+b)V−c−aT2V−1=AVm−BVnSo by comparison⇒m=−candn=−1

The vanderwall equation of a gas is $(\begin{matrix} P + \frac{a T^{2}}{V} \end{matrix}) V^{c} = (R T + b)$ . Where $a, b, c$ and $R$ are constants. If the isotherm is represented by $P = A V^{m} - B V^{n}$ , where $A$ and $B$ depends on temperature :

The correct option is A (A) $m = - c, n = - 1$
The van der waal equation of a gas,
$P + \frac{a T^{2}}{V} = (R T + b) V^{- c}$

Rearranging gives
$\Rightarrow P = (R T + b) V^{- c} - a T^{2} V^{- 1} = A V^{m} - B V^{n}$

So by comparison
$\Rightarrow m = - c a n d n = - 1$