The equation of state of a gas is $(P + \frac{a T^{2}}{V}) \times V_{1}^{c} = (R T + b)$ where a, b, c and R are constants. The isotherms can be represented by $P = A V^{m} - B V^{n}$ where A and B depend only on temperature. Then,

m = - c, n = -1

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m = c, n = 1

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m = -c, n = 1

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m = c, n = -1

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Solution

The correct option is A
m = - c, n = -1

Expanding the equation of state we have
$P V^{c} + a T^{2} V^{c - 1} = R T + b o r P = - a T^{2} V^{- 1} + R T V^{- c} + b V^{- c} o r P = A V^{- c} - B V^{- 1} . . . . . (i) w h e r e A = R T + b a n d B = a T^{2} . W e a r e g i v e n t h a t P = A V^{m} - B V^{n} . . . . . . . (i i)$
Comparing the power of V in (i) and (ii) we get m = -c and n = -1.

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The equation of state of a gas is $(P + \frac{a T^{2}}{V}) \times V_{1}^{c} = (R T + b)$ where a, b, c and R are constants. The isotherms can be represented by $P = A V^{m} - B V^{n}$ where A and B depend only on temperature. Then,

Q. 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 equation of state of a gas is given by $(P + \frac{a T^{2}}{V}) V^{c} = (R T + b)$ , where a, b, c and R are constants. The isotherms can be represented by $P = A V^{m} - B V^{n}$ , where A and B depend only on temperature and