Respuesta :
Answer :
(a) The volume occupied by 8.2 mmol of the gas under these conditions is, 0.0087 L.
(b) The value of the second virial coefficient 'b' at 300 K is -0.16
Explanation :
Part (a) :
The formula for compression factor of a gas (Z) is:
[tex]Z=\frac{PV_{real}}{nRT}[/tex]
where,
Z = impressibility factor = 0.86
P = pressure of gas = 20 atm
[tex]V_{real}[/tex] = volume of gas = ?
T = temperature of gas = 300 K
n = number of moles of gas = 8.2 mmol = 0.0082 mol (1 mmol = 0.001 mol)
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
[tex]0.86=\frac{(20atm)\times V_{real}}{(0.0082mol)\times (0.0821L.atm/mol.K)\times (300K)}[/tex]
[tex]V_{real}=0.0087L[/tex]
Thus, the volume occupied by 8.2 mmol of the gas under these conditions is, 0.0087 L.
Part (b) :
First we have to calculate the volume for ideal gas.
Formula used :
[tex]Z=\frac{V_{real}}{V_{ideal}}[/tex]
[tex]0.86=\frac{0.0087L}{V_{ideal}}[/tex]
[tex]V_{ideal}=0.010L[/tex]
Now we have to calculate the value of the second virial coefficient 'b' at 300 K.
[tex]V_{ideal}=V_{real}-n\times b[/tex]
[tex]0.010=0.0087-(0.0082\times b)[/tex]
[tex]b=-0.16[/tex]
Thus, the value of the second virial coefficient 'b' at 300 K is, -0.16
