Question

Pressure depends on distance as, $P=\frac{\alpha }{\beta }exp(-\frac{\alpha z}{k\theta })$, where $\alpha ,\beta$ are constants, z is distance, k is Boltzmann's constant and $\theta$ is temperature. The dimension of $\beta$ are.

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

Answer 1

Answer: (C)

$M^0{L}^2{T}^0$

Given,

$P=\frac{\alpha }{\beta }{e}^{\frac{-\alpha z}{k\theta }}$

Since, the exponentials are devoid of dimensions, the exponential part of the equation is ignored.

Rest we have, $P=\frac{\alpha }{\beta }$

Since, $\frac{\alpha z}{k\theta }=Dimensionless$

$\alpha =\frac{k\theta }{z}$

Kinetic energy $=\frac{3}{2}kT$

$k=\frac{K.E}{T}$

$⟹\left[k\right]=\left[M^1{L}^2{T}^{-2}\right]\left[K^{-1}\right]$

$⟹\left[z\right]=\left[L^{-1}\right]$

$⟹\left[\theta \right]=\left[K^{-1}\right]$

From these, we get the values of $\alpha$ as,

$\left[\alpha \right]=\left[M^1{L}^1{T}^{-2}\right]$

Now, we know the dimension of prressure,

$\left[P\right]=M^1{l}^{-1}{t}^{-2}]$

$\beta =\frac{\alpha }{P}$

$⟹\beta =\frac{\left[M^1{L}^1{T}^{-2}\right]}{\left[M^1{L}^{-1}{T}^{-2}\right]}$

$⟹\beta =\left[M^0{L}^2{T}^0\right]$