Solveeit Logo
Login

Question

Question: P = \(\frac{\alpha}{\beta}\)exp\(\left( - \frac{\alpha z}{K_{B}\theta} \right)\) θ → Temperature P...

P = αβ\frac{\alpha}{\beta}exp(−αzKBθ)\left( - \frac{\alpha z}{K_{B}\theta} \right)

θ → Temperature

P → Pressure

KB→ Boltzmann constant

z→ Distance

Dimension of β is –

A

M0 L0 T0M^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{0}

B

 M-1L1T2\text{ }\text{M}^{\text{-1}}L^{1}T^{2}

C

M0L2T0M^{0}L^{2}T^{0}

D

ML-1T-2\text{M}\text{L}^{\text{-1}}T^{\text{-2}}

Answer

M0L2T0M^{0}L^{2}T^{0}

Explanation

Solution

[P] = [α][β]\frac{\lbrack\alpha\rbrack}{\lbrack\beta\rbrack} ; [αz] = [KBθ] = [PV]

⇒ [α] = MLT2\Rightarrow \ \lbrack\alpha\rbrack\ = \ ΜLΤ^{2}

∴ [β] = [α][P]\frac{\lbrack\alpha\rbrack}{\lbrack P\rbrack} = L2