What is the pressure of 2 , mol of NH_3 at 27 | vedclass

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(B) The van der Waals equation is given by: $\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$ Given: $n = 2 \, mol$,$V = 5 \, L$,$T = 27 + 273 = 3

Vedclass · Accepted Answer

$9.33$

Vedclass · Answer

(B) The van der Waals equation is given by: $\left( P + \frac{n^2 a}{V^2} ight) (V - nb) = nRT$ Given: $n = 2 \, mol$,$V = 5 \, L$,$T = 27 + 273 = 300 \, K$,$a = 4.17$,$b = 0.03711$,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$. Substituting the values: $\left( P + \frac{2^2 	imes 4.17}{5^2} ight) (5 - 2 	imes 0.03711) = 2 	imes 0.0821 	imes 300$ $\left( P + \frac{16.68}{25} ight) (5 - 0.07422) = 49.26$ $(P + 0.6672) 	imes 4.92578 = 49.26$ $P + 0.6672 = \frac{49.26}{4.92578} \approx 10.00$ $P = 10.00 - 0.6672 = 9.3328 \, atm \approx 9.33 \, atm$.

What is the pressure of $2 \, mol$ of $NH_3$ at $27 \, ^oC$ when its volume is $5 \, L$ using the van der Waals equation (in $, atm$)? ($a = 4.17 \, L^2 \, bar \, mol^{-2}$,$b = 0.03711 \, L \, mol^{-1}$)

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