Question

What is the pressure of 2 mole of $N{H}_3$ 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)

T = 27 + 273 = 300 K.
R = 0.082 L atm $K^{-1}mo{l}^{-1}$
n = 2 mol, a = 4.17
V = 5 L b = 0.03711
$\left(P+\frac{a{n}^2}{V^2}\right)$ (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