Question

Pressure depends on the distance as $P = \dfrac{\alpha }{\beta }\exp \left( { - \dfrac{{\alpha z}}{{k\theta }}} \right)$. Where $\alpha $, $\beta $ are constants. $z$ is distance, $k$ is Boltzmann’s constant and $\theta $ is temperature. The dimensions of $\beta $ are

Answer 1

Hint: To find the dimensional formula of the quantity $\beta $, we need to first know the dimensional formulae for the individual quantities in the given formula. Then, we can substitute the dimensional formulae of these individual quantities in the formula.

Complete answer:
They’ve given the formula for Pressure as
$P = \dfrac{\alpha }{\beta }\exp \left( { - \dfrac{{\alpha z}}{{k\theta }}} \right)$
Where,
$\alpha $, $\beta $ are constants
$z$ is distance
$k$ is Boltzmann’s constant
$\theta $ is temperature
Now, here we have to observe that the exponential does not have any units i.e.,
$\dfrac{{\alpha z}}{{k\theta }} = \left[ {{M^0}{L^0}{T^0}} \right]$
We have the individual dimensional formulas for the quantities as
$\eqalign{
  & z = \left[ {{M^0}{L^1}{T^0}} \right] \cr
  & k = \left[ {{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}} \right] \cr
  & \theta = \left[ {{M^0}{L^0}{T^0}{K^1}} \right] \cr} $
Substituting these quantities in the above formula,
$\eqalign{
  & \dfrac{{\alpha z}}{{k\theta }} = \left[ {{M^0}{L^0}{T^0}} \right] \cr
  & \Rightarrow \dfrac{{\alpha \left[ {{M^0}{L^1}{T^0}} \right]}}{{\left[ {{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}} \right]\left[ {{M^0}{L^0}{T^0}{K^1}} \right]}} = \left[ {{M^0}{L^0}{T^0}} \right] \cr
  & \Rightarrow \dfrac{{\alpha \left[ {{M^0}{L^1}{T^0}} \right]}}{{\left[ {{M^1}{L^2}{T^{ - 2}}{K^0}} \right]}} = \left[ {{M^0}{L^0}{T^0}} \right] \cr
  & \Rightarrow \alpha \left[ {{M^{ - 1}}{L^{ - 1}}{T^2}} \right] \cr} $
Now, writing the dimension formula for the pressure we have,
$\eqalign{
  & P = \dfrac{\alpha }{\beta }\exp \left( { - \dfrac{{\alpha z}}{{k\theta }}} \right) \cr
  & \Rightarrow \left[ P \right] = \dfrac{{\left[ \alpha \right]}}{{\left[ \beta \right]}} \cr
  & \Rightarrow \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right] = \dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]}}{{\left[ \beta \right]}} \cr
  & \Rightarrow \left[ \beta \right] = \dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]}}{{\left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right]}} \cr
  & \Rightarrow \left[ \beta \right] = \left[ {{M^0}{L^2}{T^0}} \right] \cr
  & \therefore \left[ \beta \right] = \left[ {{L^2}} \right] \cr} $
Therefore, the quantity $\beta $ has the dimensions $\left[ {{L^2}} \right]$

Additional Information:
Dimensional analysis is based on the fundamental and derived quantities in physics. That is the derived physical quantities are written in the powers of the fundamental physical quantities. The fundamental physical quantities and their dimensional formulas are:

Fundamental quantity	Units in SI system	Dimensional formula
Mass	kilogram	$\left[ M \right]$
Length	meter	$\left[ L \right]$
Time	second	$\left[ T \right]$
Temperature	Kelvin	$\left[ K \right]$
Current	Ampere	$\left[ I \right]$
Amount of Substance	Moles	$\left[ N \right]$
Luminous Intensity	Candela	$\left[ J \right]$

Note:
If you find it hard to remember the dimension formula for pressure. Try to derive it from the simple formula for pressure like these
$P = \dfrac{F}{A}$
Similarly, you can derive the dimensional formula for the Boltzmann constant, which can be remembered as an average kinetic energy gas equation from the formula
$K{E_{avg}} = \dfrac{3}{2}kT$
You can also solve the problem if you have a good idea of the units of the quantities mentioned in the problem.