Question

Find out the units and dimensions of the constants a and b in the van der Waals’ equation (P+ a/V2)(V-b) = RT. [Hint: dimension of P= dimension a/V2)and dimension of b=dimension of V]
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Answer 1

ANSWER:

• Dimension of the constant a = [ M L⁵ T⁻² ].

• Dimension of the constant b = [ L³ ].

GIVEN:

• (P + a/V²)(V - b) = RT.

TO FIND:

• Dimensions of the constants a and b.

EXPLANATION:

$\boxed{ \bold{ \large{ \gray{ \left(P + \dfrac{a}{V ^{2}} \right)(V - b) = RT}}}}$

In the Vander Waals’ equation P and a/V² are added with each other, so they will have the same dimension. [ Quantities only with same dimensions can be added ]

$\boxed{\bold{\large{\gray{Pressure = \dfrac{Force}{Area}}}}}$

$\boxed{\bold{\large{\gray{Force = ma}}}}$

$\sf [\ F\ ] = [\ M\ L\ T^{-2}\ ]$

$\sf [\ A\ ]=[\ L^2\ ]$

$\sf[\ P\ ]=\dfrac{[\ M\ L\ T^{-2}\ ]}{[\ L^2\ ]}$

$\sf[\ P\ ]=[\ M\ L^{ - 1}\ T^{-2}\ ]$

$\sf [\ V^2\ ] = [\ L^3\ ]^2 = [\ L^6\ ]$

$\sf [\ P\ ] = \dfrac{[\ a\ ]}{[\ V^2\ ]}$

$\sf [\ P\ ][\ V^2\ ] = [\ a\ ]$

$\sf [\ a\ ] = [\ L^6\ ][\ M\ L^{ - 1}\ T^{-2}\ ]$

$\sf [\ a\ ] = [\ M\ L^{ 5}\ T^{-2}\ ]$

In the Vander Waals’ equation b is subtracted from V, so they will have the same dimension. [ Quantities only with same dimensions can be subtracted ]

$\sf [\ V\ ] = [\ L^3\ ]$

$\sf [\ V\ ] = [\ b\ ]$

$\sf [\ V\ ] =[\ b\ ] = [\ L^3\ ]$

Hence the dimension of [a] = [ M L⁵ T⁻² ] and of [b] = [ L³ ].