Calculate the pressure exerted by one mole of ${C O}_{2}$ gas at 273 K, if the Vander Waals constant a = 3.592 ${d m}^{6}$ atm ${m o l}^{- 2}$ . Assume that the volume occupied by ${C O}_{2}$ molecules is negligible.

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Solution

According to van der Waals equation of state for 1 mole of a gas,
P + $(\frac{a}{V^{2}})$ =(V-b) = 0
Since the volume occupied by $C O_{2}$ molecules is negligible, b = 0
So,
(P + $\frac{a}{V^{2}})$ V = RT
or,
P = $\frac{R T}{V^{2}}$ - $\frac{a}{V^{2}}$
= $\frac{0.082 \times 273}{22.4}$ - $\frac{3.592}{({22.4}^{2})}$
= 0.999375 - 0.0071588
=0.9922 atm

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Calculate the pressure exerted by one mole of CO2 gas at 273 K, if the Vander Waals constant a = 3.592 dm6 atm mol−2. Assume that the volume occupied by CO2 molecules is negligible.

According to van der Waals equation of state for 1 mole of a gas,P + (aV2) =(V-b) = 0Since the volume occupied by CO2 molecules is negligible, b = 0So, (P + aV2) V = RTor,P = RTV2 - aV2= 0.082×27322.4 - 3.592(22.42)= 0.999375 - 0.0071588=0.9922 atm

Calculate the pressure exerted by one mole of ${C O}_{2}$ gas at 273 K, if the Vander Waals constant a = 3.592 ${d m}^{6}$ atm ${m o l}^{- 2}$ . Assume that the volume occupied by ${C O}_{2}$ molecules is negligible.