Using van der wall’s equation, calculate the constant when 2 moles of a gas confined in a 4L flask exerts a pressure of 11.0 atm at a temperature of 300K. The value of b is 0.05 $Lmo{{l}^{-1}}$.

Hint: Van der Waals equation is an equation which describes the relationship between the pressure, volume, temperature and amount of real gases. It was derived by Johannes Diderik van der Waals in 1873.

Complete step by step solution:
Van der waals equation is generally a modified version of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collisions. However this law fails to explain the behavior of real gases. Therefore Van der Waals equation was devised and it helps us define the physical state of a real gas.
The Van der Waal equation defines the relationship between pressure, volume, temperature and amount of real gases. The equation can be represented as:
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$
Where P represents pressure, V represents volume, T = Temperature, R = Constant have fixed value 0.0821, n = number of moles, a and b are van der waals constant.
By putting the given values in the equation we get:
$(11+\dfrac{a{{(2)}^{2}}}{{{(4)}^{2}}})(4-2\times 0.05)=2\times 0.0821\times 300$
$(11+\dfrac{a}{4})(4-0.1)=49.26$
$\therefore a=6.523c{{m}^{3}}$
When 2 moles of a gas confined in a 4L flask exerts a pressure of 11.0 atm at a temperature of 300K then value of b is 0.05 $Lmo{{l}^{-1}}$ and constant is $6.523c{{m}^{3}}$.

Note: Van der waal equation is able to find that the behavior of gases is better than the ideal gas equation; it is not only applicable to gases but for all fluids too. But this equation gives more accurate results of all real gases only above critical temperature.