SATHEE: States of Matter - Result Question 36

States of Matter - Result Question 36

####36. For real gases, van der Waals equation is written as

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

where ’ $a$ ’ and ’ $b$ ’ are van der Waals constants.

Two sets of gases are :

(I) $O_{2}, C O_{2}, H_{2}$ and $H e$ (II) $C H_{4}, O_{2}$ and $H_{2}$ The gases given in set-I in increasing order of ’ $b$ ’ and gases given in set-II in decreasing order of ’ $a$ ‘, are arranged below. Select the correct order from the following :

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(a) (I) $H e < H_{2} < C O_{2} < O_{2}$ (II) $C H_{4} > H_{2} > O_{2}$

(b) (I) $O_{2} < H e < H_{2} < C O_{2}$ (II) $H_{2} > O_{2} > C H_{4}$

(d) (I) $H_{2} < O_{2} < H e < C O_{2}$ (II) $O_{2} > C H_{4} > H_{2}$

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Solution:

(c) Set-I : $O_{2}, C O_{2}, H_{2}, H e$ for ‘b’ values Set-II : $C H_{4}, O_{2}, H_{2},$ for ‘a’ values Value of van der Waals constant ‘b’ increases with increase in molecular volume. Clearly, the increasing order of molecular volume (size of molecule) is:

$H e < H_{2} < O_{2} < C O_{2}$

Value of ‘a’ increases with increase in intermolecular attraction. $C H_{4}$ is a polar molecule, thus, it will possess highest value of ’ $a$ ‘. $O_{2}$ and $H_{2}$ , both are non-polar molecules. The van der Waals force $\propto$ molecular mass. Hence, the correct order of value of ‘b’ is : $C H_{4} > O_{2} > H_{2}$ . From the given options, (c) is most suitable.

States of Matter - Result Question 36

Solution:

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States of Matter - Result Question 36

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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