A van der Waal's gas obeys the equation of state $(P + \frac{n^{2} a}{V^{2}}) (V - n b) = n R T$ . Its internal energy varies as $U = C T - \frac{n^{2} a}{V}$ . The equation of a quasistatic adiabat for this gas is given by-

T^{(\frac{C}{n R})} V

= constant

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T^{(\frac{C + n R}{n R})} V

= constant

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T^{(\frac{C}{n R})} (V - n b)

= constant

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P^{(\frac{C + n R}{n R})} (V - n b)

= constant

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Solution

The correct option is C $T^{(\frac{C}{n R})} (V - n b)$ = constant
For adiabatic process
$Δ Q = 0 and - Δ U = Δ W \Rightarrow - n C_{v} Δ T = P Δ V (∵ Δ W = P Δ V) when change is very small \Rightarrow - n C_{v} d T = P d V$

Now given
$U = C T - \frac{n^{2} a}{V} ∴ d U = C d T + \frac{n^{2} a}{V^{2}} d V$
put this value of $d U$ in $- d U = d W$
$∴ - (C d T + \frac{n^{2} a}{V^{2}} d V) = P d V . . . . . . (1)$

From van der waals equation
$P = (\frac{n R T}{V - n b}) - \frac{n^{2} a}{V^{2}}$
replace it in $(1)$
$- (C d T + \frac{n^{2} a}{V^{2}} d V) = ((\frac{n R T}{V - n b}) - \frac{n^{2} a}{V^{2}}) d V$
$∴ - C d T = (\frac{n R T}{V - n b}) d V$
$∴ - \frac{C}{n R} \frac{d T}{T} = \frac{d V}{V - n b}$
Integrating we get
$- \frac{C}{n R} l n T = l n (V - n b) + k$
$- l n T^{(\frac{C}{n R})} = l n (V - n b) + k$
(k $\to$ constant of integration)
$∴ l n (T^{(\frac{C}{n R})} (V - n b)) = - k$
or $T^{(\frac{C}{n R})} (V - n b)$ = constant

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Similar questions

(p + \frac{n^{2} a}{V^{2}}) (V - n b) = n R T

a

[P + \frac{n^{2} a}{V^{2}}] (V - n b) = n R T

[P + \frac{n^{2} a}{V^{2}}] (V - n b) = n R T

a

p = \frac{n R T}{V - n b} - a (\frac{n}{b})^{2}

p = \frac{n R T}{V}

P = \frac{n R T}{V - n b} - a (\frac{n}{V})^{2}

P = \frac{n R T}{V}

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