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The Van der Waals equation of state for some gases can be expressed as: $\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$, where $P$ is the pressure, $V$ is the molar volume, and $T$ is the absolute temperature of the given sample of gas and $a$, $b$ and $R$ are constants. The dimensions of $a$ are:(A) $M{L^5}{T^{ - 2}}$(B) $M{L^{ - 1}}{T^{ - 2}}$(C) ${L^3}$(D) none of the above

Last updated date: 12th Jul 2024
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Hint: By considering the other terms as the constant than the $\left( {P + \dfrac{a}{{{V^2}}}} \right)$. By using this term only, the dimension of the $a$ can be determined. By keeping the $a$ in one side and the other terms in the other side, the dimension of $a$ can be determined.

Complete step by step solution
Given that,
The Van der Waals equation of state for some gases can be expressed as: $\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$, where $P$ is the pressure, $V$ is the molar volume, and $T$ is the absolute temperature of the given sample of gas.
By considering the term,
$\left( {P + \dfrac{a}{{{V^2}}}} \right) = 0$
By rearranging the terms, then the above equation is written as,
$\left| P \right| = \left| {\dfrac{a}{{{V^2}}}} \right|$
By keeping the term $a$ in one side and the other terms in other side, then the above equation is written as,
$a = P \times {V^2}\,................\left( 1 \right)$
Now, the dimensional formula of each terms is,
The dimension of the pressure is given as,
$P = \dfrac{F}{A} = \dfrac{{ma}}{A}$
The unit of the above equation is written as,
$P = \dfrac{{kgm{s^{ - 2}}}}{{{m^2}}}$
By substituting the dimension in the above equation, then
$P = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}}$
The dimensional formula of the volume is given by,
$V = {V^2}$
The unit of the above equation is written as,
$V = {\left( {{m^3}} \right)^2}$
Then the above equation is written as,
$V = {m^6}$
By substituting the dimension in the above equation, then
$V = {L^6}$
By substituting the dimensional formula in the equation (1), then the equation is written as,
$a = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} \times {L^6}$
By rearranging the terms, then the above equation is written as,
$a = ML{T^{ - 2}} \times {L^{ - 2}} \times {L^6}$
On further simplification of the power, then
$a = M{L^5}{T^{ - 2}}$

Hence, the option (A) is the correct answer.

Note: Here the dimension of $a$ is asked, so that the term $\left( {P + \dfrac{a}{{{V^2}}}} \right)$ is taken. If the dimension of the $b$ is asked, then this term $\left( {V - b} \right)$ is taken and the solution is done like we discussed the step by step to determine the dimension formula in the above solution.