Dimension test

Classical Mechanics Level 3

$\rho = \dfrac{RT}{V-b}e^{-\dfrac{aV}{RT}}$

In the above Ideal Gas Equation ,

$V$ is the volume.
$\rho$ is the pressure.
$T$ is the temperature of gas.
R is a physical constant called Universal Gas Constant.

Then the dimensions of the quantity $(ab)$ are same as that of

Temperature Pressure Torque Mass Volume None of these Density

1 solution

Siddharth Bhatnagar
Jun 4, 2015

I solved it this way.Notice that $aV/RT$ is dimensionless . Now also $RT$ has dimensions of energy (recall $\dfrac{3}{2} RT$ is energy of a monoatomic gas due to transational motion). This gives us the dimensions of $a$ i.e pressure $P$

According to principle of homogenity of dimensions, $b$ has dimensions of Volume $V$

So dimensionsOf(ab)=dimensionsOf(PV)=dimensionsOf(energy)

$\tau= F.r$ is dimensionally consistent with energy,so torque is the required answer

Nicely Done!

Nihar Mahajan - 6 years ago

Dimension test

1 solution

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