You visited us 11 times! Enjoying our articles? Unlock Full Access!

Ideal Gas Equation

Calculate the...

Question

Calculate the pressure exerted by $22$ g of carbon dioxide in $0.5$ $d m^{3}$ at $300$ K using:
(a) the ideal gas law and
(b) Vander Waal's equation respectively.
Given:[ $a = 3.6$ atm $l i t r e^{2} m o l^{- 2}$ , $b = 0.04$ litre $m o l^{- 1}$ , $R = 0.08$ L-atom/K-mol]

Open in App

Solution

By using the ideal gas law,
$P V = n R T$
$\Rightarrow P = \frac{W}{M} \times \frac{R T}{V} = \frac{22 \times 0.0821 \times 300}{44 \times 0.5} = 24.63 a t m$
By using Van der Waals equation:
$(P + \frac{n^{2} a}{V^{2}}) (v - n b) = n R T$
$\Rightarrow P = \frac{n R T}{V - n b} - \frac{n^{2} a}{v^{2}} = \frac{\frac{22}{44} \times 0.0821 \times 300}{0.5 - \frac{22}{44} \times 40 \times 10^{- 3}} - \frac{{(\frac{22}{44})}^{2} \times 360}{(0.5)^{2} \times 101.321}$
$= 25.66 - 3.55$
$= 22.1 a t m$ .

Suggest Corrections

Similar questions

0.5 d m^{3}

a = 360 k P a d m^{6} m o l^{- 2}

b = 40 c m^{3} m o l^{- 1}

16 g

250 m L

300 K

a = 2.253 a t m L^{2} m o l^{- 2}, b = 0.0428 L m o l^{- 1}

R - 0.0821 L a t m K^{- 1} m o l^{- 1}

g

d m^{3}

K

k P a

1 mole

0.25 litre

300 K

a = 2.25 {atm - L}^{2} / {mol}^{2}

b = 0.05 L/mol

1.32 l i t r e

48^{\circ} C

16.4 a t m

(a = 3.59 a t m l i t r e^{2} m o l^{- 1}; b = 4.27 \times 10^{- 3} l i t r e m o l^{- 1}

R = 0.0821 l i t r e - a t m K^{- 1} m o l^{- 1})

Join BYJU'S Learning Program