Respuesta :
Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.
Explanation :
First we have to calculate the pressure of gas by using ideal gas equation.
[tex]PV=nRT[/tex]
where,
P = Pressure of [tex]Xe[/tex] gas = ?
V = Volume of [tex]Xe[/tex] gas = 0.805 L
n = number of moles [tex]Xe[/tex] = 1.93 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]Xe[/tex] gas = 306 K
Now put all the given values in above equation, we get:
[tex]P\times 0.805L=1.93mole\times (0.0821L.atm/mol.K)\times 306K[/tex]
[tex]P=60.2atm[/tex]
Now we have to calculate the pressure of gas by using van der Waals equation.
[tex](P+\frac{an^2}{V^2})(V-nb)=nRT[/tex]
P = Pressure of [tex]Xe[/tex] gas = ?
V = Volume of [tex]Xe[/tex] gas = 0.805 L
n = number of moles [tex]Xe[/tex] = 1.93 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]Xe[/tex] gas = 306 K
a = pressure constant = [tex]4.19L^2atm/mol^2[/tex]
b = volume constant = [tex]5.11\times 10^{-2}L/mol[/tex]
Now put all the given values in above equation, we get:
[tex](P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K[/tex]
[tex]P=44.6atm[/tex]
Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.
