Question

If the slope of ′ $ Z $ ′ (compressibility factor) v/s ′ $ P $ ′ curve is constant (slope = $ \dfrac{\pi }{{492.6}}at{m^{ - 1}} $ ) at a particular temperature ( $ 300K $ ) and very high pressure, then calculate diameter of the molecules.
(Given: $ {N_A} = 6.0 \times {10^{23}} $ , $ R = 0.0821atm \cdot litmo{l^{ - 1}}{K^{ - 1}} $ )
(A) $ {\text{7}}{\text{.5}}\mathop {\text{A}}\limits^{\text{o}} $
(B) $ {\text{5}}\mathop {\text{A}}\limits^{\text{o}} $
(C) $ {\text{2}}{\text{.5}}\mathop {\text{A}}\limits^{\text{o}} $
(D) $ {\text{1}}{\text{.25}}\mathop {\text{A}}\limits^{\text{o}} $

Answer 1

Hint: The general equation for a straight line is $ y = mx + c $ . This equation is compared to the equation relating compressibility factor $ \left( Z \right) $ with pressure $ \left( P \right) $ .

Formula used: The formulae used in the solution are given here.
For ideal gases, $ Z = 1 + \dfrac{{Pb}}{{RT}} $ where $ Z $ is the compressibility factor, $ P $ is the pressure of an ideal gas, $ R $ is the gas constant and $ T $ is the temperature.

Complete Step by Step Solution
For an ideal gas,
 $ PV = nRT $ where $ P $ is the pressure of an ideal gas, $ V $ is the volume of the ideal gas, $ n $ is the number of molecules, $ R $ is the gas constant and $ T $ is the temperature.
The compressibility factor (Z), also known as the compression factor, is the ratio of the molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure for an ideal gas the compressibility factor is 1.
At high pressures, $ Z = 1 + \dfrac{{Pb}}{{RT}} $ .
The general equation for a straight line is $ y = mx + c $ where $ x $ and $ y $ are the coordinate axes, $ m $ is the slope of the line and $ c $ is the y-intercept.
Since, the curve for ′ $ Z $ ′ (compressibility factor) v/s ′ $ P $ ′ curve, slope $ m = \dfrac{b}{{RT}} $ .
Given that, the slope for the curve is constant, slope = $ m = \dfrac{b}{{RT}} = \dfrac{\pi }{{492.6}}at{m^{ - 1}} $ .
 $ \Rightarrow b = \dfrac{\pi }{{492.6}} \times 0.0821 \times 300 $ where universal gas constant $ R = 0.0821 $ and $ T = 300K $ .
 $ b = \dfrac{4}{3}\pi {r^3} \times 4{N_A} $
 $ \Rightarrow b = \dfrac{\pi }{{492.6}} \times 0.0821 \times 300 \times {10^{ - 3}} $
The radius of the molecules are given by, $ {\text{r = 2}}{\text{.5}}\mathop {\text{A}}\limits^{\text{o}} $ .
The diameter is $ {{2r = 2 \times 2}}{\text{.5 = 5}}\mathop {\text{A}}\limits^{\text{o}} $ .
Hence, the correct answer is Option B.

Note
The compressibility factor should not be confused with the compressibility (also known as coefficient of compressibility or isothermal compressibility) of a material, which is the measure of the relative volume change of a fluid or solid in response to a pressure change.
Compressibility factor values are usually obtained by calculation from equations of state (EOS), such as the equation which takes compound-specific empirical constants as input. For a gas that is a mixture of two or more pure gases (air or natural gas, for example), the gas composition must be known before compressibility can be calculated.
Alternatively, the compressibility factor for specific gases can be read from generalized compressibility charts that plot $ Z $ as a function of pressure at constant temperature.