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Question: What is the radius of \(He\) atoms if its van der Waals constant \('b'\) is \(24ml\;mo{l^{ - 1}}\)? ...

What is the radius of HeHe atoms if its van der Waals constant ′b′'b' is 24ml  mol−124ml\;mo{l^{ - 1}}?
A. 1.355Ao1.355\mathop {\text{A}}\limits^o
B. 1.314Ao1.314\mathop {\text{A}}\limits^o
C. 1.255Ao1.255\mathop {\text{A}}\limits^o
D. 0.355Ao0.355\mathop {\text{A}}\limits^o

Explanation

Solution

As we know that van der waal constants a and b, where a represents the magnitude of intermolecular forces of attraction and b represents the effective size of molecules. Higher the size of a molecule higher will be its van der waal constants.

Formula used: b=4×43πr3×NAb = 4 \times \dfrac{4}{3}\pi {r^3} \times {N_A}

Complete answer:
As we know that van der waals constants a and b are used to characterise the specific features of real gases and the significance of a is that it represents the magnitude of intermolecular forces of attraction and b represents the effective size of molecules. The van der waals equation is modified version of ideal gas equation and it is given as:
P+an2V2(V−nb)=nRTP + a\dfrac{{{n^2}}}{{{V^2}}}(V - nb) = nRT
We know that b is a measure of size of a molecule or atoms so it can be given as:
b=4×43πr3×NAb = 4 \times \dfrac{4}{3}\pi {r^3} \times {N_A}
So we are given with the value of b as 24ml  mol−124ml\;mo{l^{ - 1}}, we can convert it into metre cube so it will be24×10−6m324 \times {10^{ - 6}}{m^3} , So now we can calculate the radius of Helium atoms as:
r3=3b16π×NA{r^3} = \dfrac{{3b}}{{16\pi \times {N_A}}}
After adjusting all these values we will get:
$
\Rightarrow r = {\left( {\dfrac{{3 \times 24 \times {{10}^{ - 6}}}}{{16 \times 3.14 \times 6.022 \times {{10}^{23}}}}} \right)^{\dfrac{1}{3}}} \\
\Rightarrow r = 1.355{m^3} \\

ButthegivenoptionsareinAngstromsowecanconvertitsoitwillbeequalto But the given options are in Angstrom so we can convert it so it will be equal to1.355 \times {10^{ - 10}}morwecansayor we can say1.355\mathop {\text{A}}\limits^o $.

**Therefore the correct answer is (A).

Additional information: **
Van der waal constant ‘a’ depends on the volume, pressure as well as number of moles. Molecules with higher value of ‘a’ will have higher attraction and molecules with weakest attractive forces will have smaller ‘a’ values. The ease of liquefaction of a gas depends on the intermolecular forces of attraction which is measured in terms of van der waal constant ‘a’.

Note:
Remember that value of van der waal constant ‘b’ depends on the size of the atoms of a molecule. Higher the size, higher will be the ‘b’ value suggesting that forces acting on the molecules are larger due to increased surface area and it provides the correction for finite molecular size. It value is basically the volume of one mole of the atoms or molecules.