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  • The equation of state of a gas is <img alt="\mathrm{\left(P+\frac{a T^2}{V}\right) \times V^c=(R T+b)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28P&plus;

The equation of state of a gas is
\mathrm{\left(P+\frac{a T^2}{V}\right) \times V^c=(R T+b)}

Where a, b, c and Rare constants. The isotherms can be represented by
\mathrm{P=A V^m-B V^n}

Where A and B depend only on temperature and

Option: 1

m = − c, n = −1


Option: 2

m = c, n = 1


Option: 3

m = − c, n = 1


Option: 4

m = c, n = −1


Answers (1)

best_answer

Expanding the equation of state we have

\mathrm{\begin{aligned} & P V^c+a T^2 V^{c-1}=R T+b \\ & \text { Or } P=-a T^2 V^{-1}+R T V^{-c}+b V^{-c} \\ & \text { Or } P=A V^{-c}-B V^{-1} \end{aligned}}             ....[i]

\mathrm{\text { Where } A=R T+b \text { and } B=a T^2 \text {. We are given that }}

\mathrm{P=A V^m-B V^n}                .....[ii]

Comparing the powers of V in (i) and (ii) we get m = −c and n = −1

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Ritika Kankaria

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