Question

In the relation, $P=\frac{\alpha }{\beta }{e}^{\alpha z/k\theta }$ $P$ is pressure, $Z$ is distance, $K$ is Boltzmann constant and $\theta$ is the temperature. The dimensions of $\beta$ will be

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

Answer 1

Answer: (A)

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

The dimension of power of e is zero.

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

Thus unit of: $\alpha =\frac{k\theta }{z}$

Unit of boltzman's constant K is $K=\left[M^1{L}^2{T}^{-2}{K}^{-1}\right]$

By putting thes values we get,

Unit of $\alpha =\frac{\left[M^1{L}^2{T}^{-2}{K}^{-1}\right]\left[K^1\right]}{\left[L^1\right]}=\left[M^1{L}^1{T}^{-2}{K}^0\right]$

Therefore unit of $P=\frac{\alpha }{\beta }$

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

$\beta =\frac{\left[M^1{L}^1{T}^{-2}{K}^0\right]}{\left[M^1{L}^{-1}{T}^{-2}\right]}=\left[M^0{L}^2{T}^0\right]$