
Using van der Waals equation, $\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$ the term that accounts for intermolecular forces in van-der Waals equation for non-ideal gas is,
A.$RT$
B.$V - b$
C.$P + \dfrac{a}{{{V^2}}}$
D.$R{T^{ - 1}}$
Answer
481.8k+ views
Hint: Ideal gas equation doesn’t hold true for real gases. Hence van der Waals introduced a new equation by modifying the ideal gas equation. The extra terms in van der Waals equation are a and b.
Complete step by step answer:
Van der Waals equation for real gas is,
$\left( {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) = nRT$
Where,
P = Pressure of gas
V = Volume of gas
n = Number of moles of gas
T = Temperature of the gas in Kelvin.
R is universal gas constant. a and b are certain constants called van der Waals constants. a corresponds to intermolecular forces and b corresponds to co-volume of excluded volume of the gas.
In the question, n=1. i.e. the number of moles of gas is one. The corresponding equation is,
$\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$
In the derivation of the ideal gas equation, it is assumed that there is no force of interaction between the gas molecules. But it is not true. A molecule in the center of the container will experience forces of interaction from all directions equally. Hence these forces cancel each other. But when the molecule reaches the wall of the container it is attracted from only one side. Hence it will strike the wall with a lower pressure than expected if there is no attraction between the molecules. Therefore, it is necessary to add a certain correction factor with the term P. This factor is called internal pressure and is given by, \[\dfrac{a}{{{V^2}}}\] . Hence the term that account for intermolecular forces in van-der Waals equation for non-ideal gas is,
$\left( {P + \dfrac{a}{{{V^2}}}} \right)$.
And hence option C is correct.
Note:
The term b in van der Waals equation corresponds to excluded volume for one mole. It is equal to four times the volume of one mole of gaseous molecules. Value of b is directly proportional to size of the molecules.
Complete step by step answer:
Van der Waals equation for real gas is,
$\left( {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) = nRT$
Where,
P = Pressure of gas
V = Volume of gas
n = Number of moles of gas
T = Temperature of the gas in Kelvin.
R is universal gas constant. a and b are certain constants called van der Waals constants. a corresponds to intermolecular forces and b corresponds to co-volume of excluded volume of the gas.
In the question, n=1. i.e. the number of moles of gas is one. The corresponding equation is,
$\left( {P + \dfrac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$
In the derivation of the ideal gas equation, it is assumed that there is no force of interaction between the gas molecules. But it is not true. A molecule in the center of the container will experience forces of interaction from all directions equally. Hence these forces cancel each other. But when the molecule reaches the wall of the container it is attracted from only one side. Hence it will strike the wall with a lower pressure than expected if there is no attraction between the molecules. Therefore, it is necessary to add a certain correction factor with the term P. This factor is called internal pressure and is given by, \[\dfrac{a}{{{V^2}}}\] . Hence the term that account for intermolecular forces in van-der Waals equation for non-ideal gas is,
$\left( {P + \dfrac{a}{{{V^2}}}} \right)$.
And hence option C is correct.
Note:
The term b in van der Waals equation corresponds to excluded volume for one mole. It is equal to four times the volume of one mole of gaseous molecules. Value of b is directly proportional to size of the molecules.
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