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Standard VI

Mathematics

Dimensional Analysis

Calculate the...

Question

Calculate the pressure (as kPa) exerted by 22 g of carbon dioxide in 0.5 $d m^{3}$ at 298. 15 K using van der Waals equation.
[Given: $a = 363.76 k P a d m^{6} m o l^{- 2}$ and $b = 43.67 c m^{3} m o l^{- 1}$ ]

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Solution

As we know,

$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$

$⎛ ⎝ P + \frac{(\frac{1}{2})^{2} \times 363.76}{(0.5)^{2}} ⎞ ⎠ (0.5 - \frac{1}{2} \times \frac{42.67}{1000}) = \frac{1}{2} \times 8.3187 \times 298.15$
$P = 2225.55 \approx 2226 k P a$

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Calculate the pressure (as kPa) exerted by 22 g of carbon dioxide in 0.5 dm3 at 298. 15 K using van der Waals equation. [Given: a=363.76kPadm6mol−2 and b=43.67cm3mol−1]

As we know,(P+an2V2)(V−nb)=nRT⎛⎝P+(12)2×363.76(0.5)2⎞⎠(0.5−12×42.671000)=12×8.3187×298.15P=2225.55≈2226kPa

Calculate the pressure (as kPa) exerted by 22 g of carbon dioxide in 0.5 $d m^{3}$ at 298. 15 K using van der Waals equation.
[Given: $a = 363.76 k P a d m^{6} m o l^{- 2}$ and $b = 43.67 c m^{3} m o l^{- 1}$ ]

As we know,

$(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$

$⎛ ⎝ P + \frac{(\frac{1}{2})^{2} \times 363.76}{(0.5)^{2}} ⎞ ⎠ (0.5 - \frac{1}{2} \times \frac{42.67}{1000}) = \frac{1}{2} \times 8.3187 \times 298.15$
$P = 2225.55 \approx 2226 k P a$