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Based on the following isotherms, select correct statements.

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For H₂ at 0ºC, molecular attractive forces are weak and size effect dominates its behavior resulting that (Z>1)

For N₂ at 0ºC, attractive forces are large enough to cause negative deviation upto 150 atm thus (Z<1) after which size effect dominates thus (Z>1)

For CO₂, intermolecular attraction is large even at 40ºC resulting (Z<1) even at high pressures

For CO₂ and N₂¸two effects compensate each other at a particular pressure where, Z=1

answer is A, B, C, D.

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Detailed Solution

1) For H₂, attractive force is zero.
Hence, $p (V - b) = R T$

$p V = R T + p b$

$Z = \frac{p V}{R T} = 1 + \frac{p b}{R T}$

Then, Z > 1

2) For N₂ (in low pressure range)
P is low , V is high
Thus, (V-b)=V

$(p + \frac{a}{V^{2}}) V = R T$

$p V = R T - \frac{a}{V}$

$Z = \frac{p V}{R T} = (1 - \frac{a}{R T V})$

Thus, Z < 1

3) As in (B)
4) At a particular pressure (separately for CO₂ and N₂) ideal behavior is reached thus, Z=1

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