Pressure depends on distance as P= (α/ß)exp(-αz/( -kθ) ), where α, ß are constants, z is distance, k is Boltzmann's constant and θ is temperature. The dimensions of ß are

To find the dimensions of $\beta$ in the equation:
$P=\frac{\alpha }{\beta }exp\left(\frac{-\alpha z}{k\theta }\right)$
where $P$ is pressure, $z$ is distance, $k$ is Boltzmann's constant, and $\theta$ is temperature, we can use dimensional analysis.

First, we analyze the exponential term. The argument of an exponential function must be dimensionless. Therefore, the quantity $\frac{\alpha z}{k\theta }$ is dimensionless:
$\left[\frac{\alpha z}{k\theta }\right]=\left[M^0{L}^0{T}^0\right]$

The term $k\theta$ represents energy (since Boltzmann's constant $k$ has dimensions of energy per temperature, $[J/K]$). The dimensions of energy are:
$\left[k\theta \right]=\left[M{L}^2{T}^{-2}\right]$

The distance $z$ has the dimension of length:
$\left[z\right]=\left[L\right]$

Now we determine the dimensions of $\alpha$:
$\left[\alpha \right]=\frac{\left[k\theta \right]}{\left[z\right]}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L\right]}=\left[ML{T}^{-2}\right]$

Next, because the exponential factor is dimensionless, the dimensions of the pressure $P$ must be equal to the dimensions of $\frac{\alpha }{\beta }$:
$\left[P\right]=\left[\frac{\alpha }{\beta }\right]$

The dimensions of pressure $P$ (force per unit area) are:
$\left[P\right]=\left[M{L}^{-1}{T}^{-2}\right]$

We can now solve for the dimensions of $\beta$:
$\left[\beta \right]=\frac{\left[\alpha \right]}{\left[P\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[M{L}^{-1}{T}^{-2}\right]}=\left[L^2\right]$

Expressing this in standard $M$, $L$, $T$ base dimensions:
$\left[\beta \right]=\left[M^0{L}^2{T}^0\right]$

Therefore, the dimensions of $\beta$ are [M⁰L²T⁰].