What is the pressure of 2 mole of $NH_3$ at 27$^\circ$C when its volume is 5 lit. in van der Waals equation (a = 4.17, b = 0 .03711)
- 10.33 aun
- 9.333 atm.
- 9.74 atm
- 9.2 atm
The Correct Option is B
Solution and Explanation
R = 0.082 L atm $K^{-1} \, mol^{-1}$
n = 2 mol, a = 4.17
V = 5 L b = 0.03711
$\left(P + \frac{an^2}{V^2} \right) $ (V - nb) = nRT
$\left(P + \frac{4.17 \times 2^2}{5^2} \right) (5 -2 \times 0.3711) = 2 \times 0.0821 \times 300 $
$\left(P + \frac{4.17 \times 4}{25} \right) (5 - 0.07422) = 600 \times 0.0821$
(P + 0.67) (4-93) = 600 $\times$ 0.0821
P + 0.67 = $\frac{600 \times 0.0821}{4.93}$
P + 0.67 = 9.99
P = 9.99 - 0.67 = 9.32 atm
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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