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)

• A. 10.33 aun
• B. 9.333 atm.
• C. 9.74 atm
• D. 9.2 atm

Answer 1

Answer: (B)

9.333 atm.

Answer 2

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