Question

Calculate the pressure exerted by one mole of CO2 gas in a 1.32 dm3 vessel at 48°C using van der Waals equation. [given ; a = 3.59 dm6 atm mol-2 ; b = 0.0427 dm3 mol-1 ]​

Answer 1

Answer:

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Answer 2

Answer:

The pressure exerted by one mole of CO₂ is equal to 18.55atm.

Explanation:

We know the Vander Waal's equation for one mole real gas:

$(P+\frac{a}{V^{2}} )(V-b)=RT$                                ...............(1)

Given:

The volume of carbon dioxide gas, $V=1.32dm^{3}$

The value of constant $a=3.59dm^{6}atmmol^{-2}$

The value of $b=0.0427dn^{3}mol^{-1}$

Temperature, $T=273+48=321K$

The value of gas constant $R=0.082dm^{3}atmK^{-1}mol^{-1}$

Put all these values in equation (1):

$(P+\frac{3.59}{(1.32)^{2}} )(1.32-0.0427)=(0.082)(321)$

$(P+2.06)(1.2773)=26.322$

$P+2.06=\frac{26.322}{1.2773}$

$P+2.06=20.607\\P=18.55atm$

Therefore, the pressure of one mole of carbon dioxide is 18.55atm.