The pressure exerted by 5 , mol of a real gas | vedclass

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(B) Using the van der Waals equation: $\left( P + \frac{an^2}{V^2} \right) (V - nb) = nRT$ Given: $a = 3.592 \, atm \, L^2 \, mol^{-2}$,$b = 0.0427 \,

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$77.2$

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(B) Using the van der Waals equation: $\left( P + \frac{an^2}{V^2} ight) (V - nb) = nRT$ Given: $a = 3.592 \, atm \, L^2 \, mol^{-2}$,$b = 0.0427 \, L \, mol^{-1}$,$V = 1 \, L$,$n = 5 \, mol$,$T = 273 + 47 = 320 \, K$,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$ Substituting the values: $\left( P + \frac{3.592 	imes 5^2}{1^2} ight) (1 - 5 	imes 0.0427) = 5 	imes 0.0821 	imes 320$ $\left( P + 89.8 ight) (1 - 0.2135) = 131.36$ $(P + 89.8) (0.7865) = 131.36$ $P + 89.8 = \frac{131.36}{0.7865} \approx 167.02$ $P = 167.02 - 89.8 = 77.22 \, atm$ Thus,the pressure is approximately $77.2 \, atm$.

The pressure exerted by $5 \, mol$ of a real gas in a $1 \, L$ container at $47 \, ^oC$ is ............. $atm$. $\left( a = 3.592 \, atm \, L^2 \, mol^{-2}, b = 0.0427 \, L \, mol^{-1} \right)$

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