For one mole of Vander Waal-s gas when b -0 and T -300 K- the PV V - S 1- v plot is shown below- The value of the van Der waal-s constant a in atm litre 2 mol -2 A- 1B- 2C- 3D- 4

The correct option is C 3The Vander Waal’s equation for 1 mole of gas will be [ p+a/V^2 ]× [v b]=RT But b = 0 ∴ [ p+a/V^2 ]× v=RT PV + a/V=RT ⇒ PV = RT a/V ...

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Standard XII

Physics

Real Gas Equation

For one mole ...

Question

For one mole of Vander Waal’s gas when b = 0 and T = 300 K, the PV $V / S \frac{1}{v}$ plot is shown below. The value of the van Der waal’s constant a
(in atm $(l i t r e)^{2} m o l^{- 2})$ is

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Solution

The correct option is C 3
The Vander Waal’s equation for 1 mole of gas will be
$[p + \frac{a}{V^{2}}] \times [v - b] = R T$
But b = 0
$∴ [p + \frac{a}{V^{2}}] \times v = R T$
$P V + \frac{a}{V} = R T \Rightarrow P V = R T - \frac{a}{V}$
When Pv is plotted against $(\frac{1}{V})$ we get straight line with negative slope i.e (–a)
$S l o p e = - a = \frac{58.8 - 67.8}{3 - 0}$
$= \frac{- 9}{3} = - 3$
-a=-3
a=+3

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For one mole of Vander Waal’s gas when b = 0 and T = 300 K, the PV V/S 1v plot is shown below. The value of the van Der waal’s constant a (in atm (litre)2 mol−2) is

The correct option is C 3The Vander Waal’s equation for 1 mole of gas will be [p+aV2]×[v−b]=RT But b = 0 ∴ [p+aV2]×v=RT PV+aV=RT⇒PV=RT−aV When Pv is plotted against (1V) we get straight line with negative slope i.e (–a) Slope=−a=58.8−67.83−0 =−93=−3 -a=-3 a=+3

For one mole of Vander Waal’s gas when b = 0 and T = 300 K, the PV $V / S \frac{1}{v}$ plot is shown below. The value of the van Der waal’s constant a
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