Question 195: Thermodynamics & KTG

Question

The equation for real gas is given by $\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$, where $\mathrm{P}, \mathrm{V}, \mathrm{T}$ and R are the pressure, volume, temperature and gas constant, respectively. The dimension of $\mathrm{ab}^{-2}$ is equivalent to that of :

Accepted Answer

The equation for a real gas is given by:
$ \left(\mathrm{P} + \frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT} $
Here, $ \mathrm{P} $, $ \mathrm{V} $, $ \mathrm{T} $, and $ R $ are the pressure, volume, temperature, and gas constant, respectively.
To find the dimension of $ ab^{-2} $, we proceed with the following steps:
Rearrange the equation for dimensions:
$ [\mathrm{a}] = [\mathrm{P}][\mathrm{V}^2] $
Since $[\mathrm{P}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}$ and $[\mathrm{V}] = \mathrm{L}^3$, we have:
$ [\mathrm{a}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}(\mathrm{L}^6) = \mathrm{ML}^5\mathrm{T}^{-2} $
As derived from the equation:
$[\mathrm{b}] = [\mathrm{V}] = \mathrm{L}^3$
Now, to find $[\mathrm{ab}^{-2}]$:
$[\mathrm{ab}^{-2}] = \frac{[\mathrm{a}]}{[\mathrm{b}]^2} = \mathrm{ML}^5\mathrm{T}^{-2} \cdot \mathrm{L}^{-6} = \mathrm{ML}^{-1}\mathrm{T}^{-2}$
This dimension, $\mathrm{ML}^{-1}\mathrm{T}^{-2}$, is equivalent to the dimension of energy density.

Answer

Compressibility

Answer

Planck's constant

Answer

Energy density

Answer

Strain

Thermodynamics & KTG

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