Question

The equation $\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$ constant $P$ is the pressure, $V$ is the volume and a, b are constants. Then unit of $a$ is
$\left( A \right)dyne \times c{m^5}$
$\left( B \right)dyne \times c{m^4}$
$\left( C \right)dynec{m^{ - 3}}$
$\left( D \right)dynec{m^{ - 2}}$

Answer 1

We know that only the units with the same dimensions can be added. Apply the same logic to the above equation. Write the pressure and volume units in the CGS system, since the options are given in terms of the CGS system. From this we can determine the unit of a. In the CGS system we write in terms of centimetre, gram and second.

Complete step by step answer:
The above equation is a Van Der Waals equation. Physical state of a real gas can be defined using the Van Der Waals equation. It tells us about the molecular size and its interactive force.
In the CGS system we express gram as the unit of mass, centimetre as the unit of length, and the second as the unit of time.
We know that we cannot subtract or add quantities of different dimensions. Hence $P$ and $\dfrac{a}{{{V^2}}}$have the same unit and dimension.
Here $P$ is the Pressure and $V$ is the volume.
Pressure is the ratio of force and area.
$P = \dfrac{{dyne}}{{c{m^2}}}$
Force unit is dyne in CGS system and area unit in CGS system is $c{m^2}$
$P$ and $\dfrac{a}{{{V^2}}}$ have the same unit and dimension.
$P = \dfrac{a}{{{V^2}}}$
$a = P \times {V^2}$
$a = \dfrac{{dyne}}{{c{m^2}}} \times {\left( {c{m^3}} \right)^2}$
$a = dyne \times c{m^4}$

Hence option $\left( B \right)$ is the right option.

Note: The force which produces an acceleration of $1c{m^2}$ in a body of mass of one gram, in its direction is called dyne. The CGS system expresses gram as the unit of mass, centimetre as the unit of length, and the second as the unit of time. Force unit is dynes in CGS system.