
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
Answer
134.1k+ views
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.
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.
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