Question:
The value of van derWaals' constant a for the gases $O_2, N_2$, $NH_3$ and $ CH_4$ are 1.360, 1.390,4.170 and $2.253L^2$ atm $mol^{-2}$ respectively. The gas which can most easily be liquefied is
The value of van derWaals' constant a for the gases $O_2, N_2$, $NH_3$ and $ CH_4$ are 1.360, 1.390,4.170 and $2.253L^2$ atm $mol^{-2}$ respectively. The gas which can most easily be liquefied is
Updated On: Aug 1, 2022
- $O_2$
- $N_2$
- $NH_3$
- $CH_4$
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The Correct Option is C
Solution and Explanation
The ease of liquefication of a gas depends on their intermolecular force of attraction which in turn is measured in terms of van der Waals' constant a. Hence, higher the value of a, greater the intermolecular force of attraction, easier the liquefication.
In the present case, $NH_3$ has highest a, can most easily be liquefied.
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Concepts Used:
Van Der Waals Equation
Van der Waals equation is an equation relating the relationship between the pressure, volume, temperature, and amount of real gases.
Read More: Derivation of Van Der Waals Equation
Derivation of Van der Waals equation:
For a real gas containing ‘n’ moles, the equation is written as
Where, P, V, T, n are the pressure, volume, temperature and moles of the gas. ‘a’ and ‘b’ constants specific to each gas.
Where,
Vm: molar volume of the gas
R: universal gas constant
T: temperature
P: pressure
V: volume
Thus, Van der Waals equation can be reduced to ideal gas law as PVm = RT.
The equation can further be written as;
- Cube power of volume:
- Reduced equation (Law of corresponding states) in terms of critical constants:
Units of Van der Waals equation Constants
a: atm lit² mol-²
b: litre mol-¹