Question

A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the ideal gas law and the van der Waals equation (van der Waals constants for Xe are a = 4.19 L2atm/mol2 and b = 5.11×10-2 L/mol).

Answer 1

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.

`PV=nRT`

where,

P = Pressure of
`Xe` gas = ?

V = Volume of
`Xe` gas = 0.805 L

n = number of moles
`Xe` = 1.93 mole

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`Xe` gas = 306 K

Now put all the given values in above equation, we get:

`P* 0.805L=1.93mole* (0.0821L.atm/mol.K)* 306K`

`P=60.2atm`

Now we have to calculate the pressure of gas by using van der Waals equation.

`(P+(an^2)/(V^2))(V-nb)=nRT`

P = Pressure of
`Xe` gas = ?

V = Volume of
`Xe` gas = 0.805 L

n = number of moles
`Xe` = 1.93 mole

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`Xe` gas = 306 K

a = pressure constant =
`4.19L^2atm/mol^2`

b = volume constant =
`5.11* 10^(-2)L/mol`

Now put all the given values in above equation, we get:

`(P+((4.19L^2atm/mol^2)* (1.93mole)^2)/((0.805L)^2))[0.805L-(1.93mole)* (5.11* 10^(-2)L/mol)]=1.93mole* (0.0821L.atm/mol.K)* 306K`

`P=44.6atm`

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.