States of Matter - Result Question 98
####2. Consider the following table.
| $\mathbf{G a s}$ | $\boldsymbol{a} /\left(\mathbf{( k ~ P a ~ d m} mol^{-\mathbf{1}}\right)$ | $\boldsymbol{b} /\left(dm^{\mathbf{3}} mol^{\mathbf{- 1}}\right)$ |
|---|---|---|
| $A$ | 642.32 | 0.05196 |
| $B$ | 155.21 | 0.04136 |
| $C$ | 431.91 | 0.05196 |
| $D$ | 155.21 | 0.4382 |
$a$ and $b$ are van der Waals’ constants. The correct statement about the gases is
(2019 Main, 10 April I)
(a) gas $C$ will occupy lesser volume than gas $A$; gas $B$ will be lesser compressible than gas $D$
(b) gas $C$ will occupy more volume than gas $A$; gas $B$ will be more compressible than gas $D$
(c) gas $C$ will occupy more volume than gas $A$; gas $B$ will be lesser compressible than gas $D$
(d) gas $C$ will occupy more volume than gas $A$; gas $B$ will be lesser compressible than gas $D$
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Answer:
Correct Answer: 2. (c)
Solution:
- For 1 mole of a real gas, the van der Waals’ equation is
$$ \left(p+\frac{a}{V^{2}}\right)(V-b)=R T $$
The constant ’ $a$ ’ measures the intermolecular force of attraction of gas molecules and the constant ’ $b$ ’ measures the volume correction by gas molecules after a perfectly inelastic binary collision of gas molecules.
For gas $A$ and gas $C$ given value of ’ $b$ ’ is
$0.05196 dm^{3} mol^{-1}$. Here,
$a \propto$ intermolecular force of attraction
$\propto$ compressibility $\propto$ real nature
$\propto \frac{1}{\text { volume occupied }}$
Value of $a /\left(kPa dm^{6} mol^{-1}\right)$ for gas $A(642.32)>\operatorname{gas} C(431.91)$ So, gas $C$ will occupy more volume than gas $A$. Similarly, for a given value of $a$ say $155.21 kPa dm^{6} mol^{-1}$ for gas $B$ and gas $D$
$\frac{1}{b} \propto$ intermolecular force of attraction
$\propto$ compressibility $\propto$ real nature
$\propto \frac{1}{\text { volume accupied }}$
$b /\left(dm^{3} mol^{-1}\right)$ for gas $B(0.04136)<\operatorname{Gas} D(0.4382)$
So, gas $B$ will be more compressible than gas $D$.
