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Question: Unit of \(\alpha \) is \(atm{L^2}mo{l^{ - 2}}\). If true enter \({\text{1}}\), else enter \({\text...

Unit of α\alpha is atmL2mol−2atm{L^2}mo{l^{ - 2}}.
If true enter 1{\text{1}}, else enter 0{\text{0}}.
A) 1{\text{1}}

Explanation

Solution

The term given as α\alpha denotes the Vander waals constant which is related to the average force of attraction between the molecules. This is related to the non-ideal gases and its equation. One can use this information to find out the unit of the given term and see if it is true or false.

Complete step by step answer:

  1. First of all as the only term and unit is given we need to find out the actual unit for this term and can find out that by using the Vander waals equation.
  2. The term ideal gas law elaborates as the molecules of a gas as the point particles which shows the perfect elastic collisions and this concept is used for the dilute gases in case of many experimental situations.
  3. As the gas molecules are not just the point masses and there are many instances where the gas is needed to be treated as the non-ideal gas. Hence taking this into consideration a modification for this is done taking the molecular size and molecular interaction forces for its calculation. The formed equation is referred to as the Vander Waals equation which is as follows,
    RT=[P+α(nV2)](Vn−b)RT = \dfrac{{[P + \alpha \left( {\dfrac{n}{{{V^2}}}} \right)]}}{{\left( {\dfrac{V}{{n - b}}} \right)}}
  4. The constant α\alpha provides a value for the intermolecular forces. Where the constant b denotes for the value occupied by the particles of gas. This value represents the correction for a molecular size which is finite and the value is dependent on the volume of one mole of the atoms or molecules and the unit is atmL2mol−2atm{L^2}mo{l^{ - 2}}.
  5. Hence, the Unit of α\alpha is atmL2mol−2atm{L^2}mo{l^{ - 2}} which is true.

Note:
In the above equation the constants α\alpha and b are called van der Waals constants. They possess positive values and they represent the characteristic for the individual gas. In case of when a gas behaves ideally the values, both α\alpha and b are zero and the van der Waals equations for an ideal gas will become as per the ideal gas law which is PV=nRTPV = nRT.