Question

In the equation $P = \dfrac{\alpha }{\beta }{e^{\dfrac{{\alpha x}}{{nR\theta }}}}$ , where $P$ is power, $x$ is distance, $n$ is number of moles, $R$ is gas constant and $\theta$ is temperature. The dimensional formula of $\beta$ is
A. $[{M^0}{L^0}{T^0}]$
B. $[{M^1}{L^0}{T^1}]$
C. $[{M^0}{L^{ - 1}}{T^1}]$
D. $[{M^0}{L^{ - 1}}{T^{ - 1}}]$

Answer 1

In this question, we have to calculate the dimensional formula of $\beta$ for the given equation. For that, we will make use of the units of the other quantities to find their dimensional formula and by solving the powers of dimensional formula, we get the desired answer.

Complete step by step answer:
The given equation is
$P = \dfrac{\alpha }{\beta }{e^{\dfrac{{\alpha x}}{{nR\theta }}}}$
Where, $P$ is power, $x$ is distance, $n$ is number of moles, $R$ is gas constant and $\theta$ is temperature.
Let us write the units of the given quantities.
$\text{Power} = \dfrac{\text{work}}{\text{time}}$ , Unit - Joule per second or watt, Dimensional formula - $[{M^1}{L^2}{T^{ - 3}}]$.
$x$ - distance , Unit – meter , Dimensional formula - $[{M^0}{L^1}{T^0}]$.
$n$ - number of moles, Unit – $mol$ , Dimensional formula - $[{M^0}{L^0}{T^0}mo{l^1}]$.
$R$ - gas constant, Unit - $Jmo{l^{ - 1}}{K^{ - 1}}$, Dimensional formula - $[{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}mo{l^{ - 1}}]$.
$\theta$ - temperature, Unit – Kelvin , Dimensional formula - $[{M^0}{L^0}{T^0}{K^1}]$.

Here, exponential term is dimensionless, so
$\left[ {\dfrac{{\alpha x}}{{nR\theta }}} \right] = 1$
$\Rightarrow \left[ \alpha \right] = \left[ {\dfrac{{nR\theta }}{x}} \right]$
Substituting dimensional formulae of the respective quantities, we get
$\left[ \alpha \right] = \left[ {\dfrac{{[{M^0}{L^0}{T^0}mo{l^1}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}mo{l^{ - 1}}][{M^0}{L^0}{T^0}{K^1}]}}{{[{M^0}{L^1}{T^0}]}}} \right]$
Solving for the powers, we have
$\left[ \alpha \right] = \left[ {{M^1}{L^1}{T^{ - 2}}} \right] - - - - - - - - - - - - - (1)$
Now, $\dfrac{\alpha }{\beta }$ must have the dimensions of $P$ Power. Hence,
$\left[ {\dfrac{\alpha }{\beta }} \right] = \left[ P \right]$
$\Rightarrow \left[ {\dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]}}{{\left[ \beta \right]}}} \right] = \left[ {[{M^1}{L^2}{T^{ - 3}}]} \right]$
$\Rightarrow \left[ \beta \right] = \left[ {\dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]}}{{\left[ {[{M^1}{L^2}{T^{ - 3}}]} \right]}}} \right]$
Solving for powers, we get
$\therefore \left[ \beta \right] = = \left[ {{M^0}{L^{ - 1}}{T^1}} \right]$

Hence, option C is correct.

Note: We should know all the units of the quantities given in the question and know about the basic dimensions of the units. Make the calculations of powers of the dimensions properly. We can also use the units of the quantities and make the calculations as required to get the desired dimensions.