Question

What is the unit of “a” in Vander Waal’s gas equation?
A.\[Atm \times litr{e^{ - 2}}mo{l^2}\]
B.\[Atm \times litr{e^{ - 2}}/mol\]
C.\[Atm \times litr{e^{ - 1}}mo{l^2}\]
D.\[Atm \times litr{e^2}mo{l^{ - 2}}\]

Answer 1

Hint: For answering this question we should learn about Van der waal’s gas equation. The Van der waal’s equation is better suited for describing the behaviour of real gas than ideal gas equation. we will understand the Van der waal’s equation and calculate the unit of ‘a’ in this question

Complete answer:
The expression for Van der waal’s equation is given as following
\[(P + \dfrac{{a{n^2}}}{{{V^2}}})(V - nb) = nRT\]
where P is the pressure of the real gas, V is the volume of the real gas, n is the number of moles of real gas, R is Universal gas constant, a and b are Van der waal’s constant
From the above equation we can observe that \[\dfrac{{a{n^2}}}{{{V^2}}}\] should have the dimension of pressure for the equation to be dimensionally correct. So, using this we will calculate the unit of Van der waal’s constant a
\[\dfrac{{a{n^2}}}{{{V^2}}} = Atm\]
Simplifying equation, we get
\[a = \dfrac{{Atm \times {V^2}}}{{{n^2}}}\]
Substituting the units in place of Volume and number of moles, we get
\[a = \dfrac{{Atm \times litr{e^2}}}{{mo{l^2}}}\]
\[a = Atm \times litr{e^2}mo{l^{ - 2}}\]
We get the required unit of Van der waal’s constant \[a = Atm \times litr{e^2}mo{l^{ - 2}}\].

Therefore, option D is the correct answer.

Note:
The Van der waal’s constant a and b are characteristic properties of a specific real gas. The constant a is responsible for the correction of force of attraction between gas particles, higher the value of ‘a’ stronger will be the force of attraction between the gas particles. The value of ‘a’ also tells us about the boiling point of the gases. The Van der waal’s constant ‘b’ is a measure of the volume, it is also known as co-volume of a gas.