At a temperature of 27,^{circ}C,if the volume | vedclass

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(B) The van der Waals equation is given by: $\left( P + \frac{n^2 a}{V^2} ight) (V - nb) = nRT$ Given: $n = 2\, 	ext{mol}$,$V = 5\, L$,$a = 4.17$,$

Vedclass · Accepted Answer

$9.333$

Vedclass · Answer

(B) The van der Waals equation is given by: $\left( P + \frac{n^2 a}{V^2} ight) (V - nb) = nRT$ Given: $n = 2\, 	ext{mol}$,$V = 5\, L$,$a = 4.17$,$b = 0.03711$,$R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$,$T = 27 + 273 = 300\, K$. Substituting the values: $\left( P + \frac{2^2 	imes 4.17}{5^2} ight) (5 - 2 	imes 0.03711) = 2 	imes 0.0821 	imes 300$ $\left( P + \frac{16.68}{25} ight) (5 - 0.07422) = 49.26$ $(P + 0.6672) 	imes 4.92578 = 49.26$ $P + 0.6672 = \frac{49.26}{4.92578} \approx 10.00$ $P = 10.00 - 0.6672 = 9.3328\, atm \approx 9.333\, atm$.

At a temperature of $27\,^{\circ}C$,if the volume of $2\, \text{moles}$ of $NH_3$ gas is $5\, L$ according to the van der Waals equation,what will be its pressure in $atm$? ($a = 4.17$,$b = 0.03711$)

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