Question

In the relation $P = \frac{\alpha}{\beta}e^{\frac{-\alpha z}{K\theta}}P$ is pressure, $Z$ is the distance, $K$ is Boltzmann's constant and $\theta$ is the temperature. The dimensional formula of $\alpha$ will be:

• A. $[M^1 L^1 T^{-2}]$
• B. $[M^1 L^2 T^{1}]$
• C. $[M^1 L^0 T^{-1}]$
• D. $[M^0 L^2 T^{-1}]$

Answer 1

Answer: (A)

$[M^1 L^1 T^{-2}]$

Answer 2

The given relation is $P = \frac{\alpha}{\beta}e^{\frac{-\alpha z}{K\theta}}$. Here, $P$ is pressure, $Z$ is the distance, $K$ is Boltzmann's constant and $\theta$ is the temperature.

For the equation to be dimensionally consistent, the exponent of the exponential term must be dimensionless. So, the dimension of $\frac{-\alpha z}{K\theta}$ is $[M^0 L^0 T^0]$. This means the dimension of $\frac{\alpha z}{K\theta}$ is also $[M^0 L^0 T^0]$.

$[\frac{\alpha z}{K\theta}] = [M^0 L^0 T^0]$

$[\alpha] [z] [K\theta]^{-1} = [M^0 L^0 T^0]$

$[\alpha] [z] = [K\theta]$

We need to find the dimensions of $[z]$ and $[K\theta]$.

$Z$ is distance, so $[z] = [L]$.

$K$ is Boltzmann's constant and $\theta$ is temperature. The product $K\theta$ has the dimensions of energy. This can be seen from the ideal gas law $PV = NKT$, where $PV$ has dimensions of energy ($[P] = [ML^{-1}T^{-2}]$, $[V] = [L^3]$, so $[PV] = [ML^2T^{-2}]$), $N$ is the number of molecules (dimensionless), so $[KT] = [PV] = [ML^2T^{-2}]$. Since the temperature is denoted by $\theta$, $[K\theta] = [ML^2T^{-2}]$.

Substitute these dimensions into the equation $[\alpha] [z] = [K\theta]$:

$[\alpha] [L] = [ML^2T^{-2}]$

Now, solve for the dimension of $\alpha$:

$[\alpha] = \frac{[ML^2T^{-2}]}{[L]}$

$[\alpha] = [ML^{2-1}T^{-2}]$

$[\alpha] = [ML^1T^{-2}]$

The dimensional formula of $\alpha$ is $[M^1 L^1 T^{-2}]$.

Therefore, the correct answer is (A).