Question

In: Chemistry

1/ A 10.64 mol sample of xenon gas is maintained in a 0.8062 L container at...

1/ A 10.64 mol sample of xenon gas is maintained in a 0.8062 L container at 299.9 K. What is the pressure in atm calculated using the van der Waals' equation for Xe gas under these conditions? For Xe, a = 4.194 L2atm/mol2 and b = 5.105×10-2 L/mol.

(......) atm

2/ According to the ideal gas law, a 0.9174 mol sample of krypton gas in a 1.363 L container at 267.1 K should exert a pressure of 14.75atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Kr gas, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol.  (......) %

3/ According to the ideal gas law, a 9.560 mol sample of argon gas in a 0.8480 L container at 495.3 K should exert a pressure of 458.2 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Ar gas, a = 1.345 L2atm/mol2 and b = 3.219×10-2 L/mol. (......) %

Solutions

Expert Solution

1.

Van Der Waal's equation of state for a real gas is given by the following equation:

(P + a*(n/V)2)*((V/n)-b) = RT

We need to put the given values in the above equation, as shown below:

(P + 4.194*(10.64/0.8062)2)*((0.8062/10.64) - 0.05105) = 0.0821*299.9

This can be simplified to:

(P + 730.51)*0.0247 = 24.6218

So,

P = 266.324 atm

2.

Using the Van Der waal's equation of state in the same way as above, we calculate the pressure in this case as:

(P + 2.318*(0.9174/1.363)2)*((1.363/0.9174) - 0.03978) = 0.0821*261.7

Calculating in same way as above:

P = 13.809 atm

So the percent difference is calculated as:

((Ideal pressure - Van der Waal pressure)/Ideal pressure)*100

Putting values:

% difference = ((14.75-13.809)/14.75)*100 = 6.38%

3.

Using the Van Der waal's equation of state in the same way as above, we calculate the pressure in this case as:

(P + 1.345*(9.56/0.8480)2)*((0.8480/9.56) - 0.03219) = 0.0821*495.3

Calculating in same way as above:

P = 548.61 atm

So the percent difference is calculated as:

((|Ideal pressure - Van der Waal pressure|)/Ideal pressure)*100

% difference = ((|458.2-548.61|)/458.2)*100 = 19.73%


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