Question

In a typical combustion engine, the work done by a gas molecule is given by

$W={\alpha }^2\beta e^{\frac{-\beta x^2}{kT}}$

where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be

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

Answer 1

The correct option is A

$\left[M^{0}L{T}^0\right]$

Step 1: Given data

Work done,

$W={\alpha }^2\beta e^{\frac{-\beta x^2}{kT}}$

We have to find the dimensions of $\alpha$

Step 2: Solution

Since, the power cannot have any dimensions, $\frac{-\beta x^2}{kT}$ has to be dimensionless.

$$\begin{gathered} \Rightarrow \left[-\beta x^2\right]=\left[kT\right] \\ \Rightarrow \left[\beta \right]=\left[\frac{kT}{x^2}\right] \end{gathered}$$

We know,

$$\begin{gathered} \left[k\right]=\left[M{L}^2{T}^{-2}{K}^{-1}\right] \\ \left[T\right]=\left[K\right] \\ \left[x\right]=\left[L\right] \\ \Rightarrow \left[\beta \right]=\frac{\left[M{L}^2{T}^{-2}{K}^{-1}\right]\left[K\right]}{{\left[L\right]}^2} \\ \Rightarrow \left[\beta \right]=\left[M{T}^{-2}\right] \end{gathered}$$

Also dimension of $LHS$ must be equal to the dimension of $RHS$

$\therefore \left[W\right]=\left[{\alpha }^2\beta e^{\frac{-\beta x^2}{kT}}\right]$

Since $\frac{-\beta x^2}{kT}$ is dimensionless,

$\therefore \left[W\right]=\left[{\alpha }^2\beta \right]$

We know,

$$\begin{gathered} \left[W\right]=\left[M{L}^2{T}^{-2}\right] \\ \left[\beta \right]=\left[M{T}^{-2}\right] \\ \Rightarrow \left[W\right]=\left[{\alpha }^2\beta \right] \\ \Rightarrow \left[{\alpha }^2\right]=\frac{\left[W\right]}{\left[\beta \right]} \\ \Rightarrow \left[{\alpha }^2\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[M{T}^{-2}\right]} \\ \Rightarrow \left[{\alpha }^2\right]=\left[L^2\right] \\ \Rightarrow \left[\alpha \right]=\left[M^{0}L{T}^0\right] \end{gathered}$$

Therefore, the dimension of $\alpha$ is $\left[M^{0}L{T}^0\right]$

Option A is correct.