Question:

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)

Updated On: Jun 18, 2022
  • 10.33 aun
  • 9.333 atm.
  • 9.74 atm
  • 9.2 atm
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The Correct Option is B

Solution and Explanation

T = 27 + 273 = 300 K.
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;

  1. Cube power of volume:
  2. 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