For real gases van der Waals equation is written as: ≤ft( P + an^2/V^2 ) (V - nb) = nRT Select the correct order for constants ' a ' and ' b ': - Chemistry MCQ Chemistry | ExamTest.live

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For real gases van der Waals equation is written as: (P + \frac{a n ^{2}}{V ^{2}}) (V - nb) = n RT Select the correct order for constants ' a' and ' b':

A $Increasing order of ' b': H_{2} < O_{2} < He < CO_{2}$
B $Decreasing order of ' a': CH_{4} > H_{2} > O_{2}$
C $Increasing order of ' b': O_{2} < He < H_{2} < CO_{2}$
D $Increasing order of ' b': H_{2} < He < O_{2} < CO_{2} and Decreasing order of ' a': CH_{4} > O_{2} > H_{2}$

Solution & Step-by-step Explanation

* Constant ' a' represents intermolecular forces of attraction. Larger/more polarizable molecules have higher ' a'. Order: CH_{4} > O_{2} > H_{2} .Constant ' b' represents the excluded volume (size of molecules). Order: H_{2} < He < O_{2} < CO_{2} .(Note: H_{2} is actually slightly smaller than He in effective volume due to bond length vs atomic radius).

Try it yourself before checking the explanation above.

For real gases van der Waals equation is written as: (P + \frac{a n ^{2}}{V ^{2}}) (V - nb) = n RT Select the correct order for constants ' a' and ' b':

Increasing order of ' b': H_{2} < O_{2} < He < CO_{2}

Decreasing order of ' a': CH_{4} > H_{2} > O_{2}

Increasing order of ' b': O_{2} < He < H_{2} < CO_{2}

Increasing order of ' b': H_{2} < He < O_{2} < CO_{2} and Decreasing order of ' a': CH_{4} > O_{2} > H_{2}

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