15.0 moles of gas are in a 6.00L tank at 298.15K. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300 L^2*atm/mol^2 and b=0.0430 L/mol.
I found the idea gas pressure to be 61.2 atm by using the ideal gas law. However I can't seem to get the right value for the pressure of methane. Below are my workings, can anyone see what I'm doing wrong?
P= [(nRT)/(V-nb)] - [(an)^2/V^2]
P=[(15 mol*0.08206*298.15K)/(6.00L-(15mol*0.0430)] -[(15mol*2.3)^2/(6.00L)^2]
P=68.53-33.06
P=35.47 atm
In my text the second term is an^2 and not (an)^2.
68.53 is ok.
The second term is -[(15)^2 x 2.3]/36 = ?
what is the absolute temperature of the gas at which 3.33 10-3 mol occupies 440. mL at 708 torr
To find the pressure of methane using the van der Waals equation, you correctly substituted the values for moles (n = 15.0 mol), ideal gas constant (R = 0.08206 L*atm/(mol*K)), temperature (T = 298.15 K), and volume (V = 6.00 L). However, there seems to be a mistake in your calculation for the term involving the van der Waals constants.
Let's calculate the pressure of methane using the van der Waals equation step by step:
1. Calculate the term nb:
nb = (15 mol) * (0.0430 L/mol) = 0.645 L
2. Calculate the term (an)^2:
(an)^2 = (15 mol) * (2.300 L^2*atm/mol^2) = 34.5 L^2*atm/mol
3. Substitute the values into the van der Waals equation:
P = [(nRT) / (V - nb)] - [(an)^2 / V^2]
P = [(15 mol * 0.08206 L*atm/(mol*K) * 298.15 K) / (6.00 L - 0.645 L)] - [34.5 L^2*atm/mol / (6.00 L)^2]
4. Evaluate the numerator:
15 mol * 0.08206 L*atm/(mol*K) * 298.15 K = 364.144 atm*L
5. Evaluate the denominator:
6.00 L - 0.645 L = 5.355 L
6. Substitute the numerator and denominator into the equation:
P = [364.144 atm*L / 5.355 L] - [34.5 L^2*atm/mol / (6.00 L)^2]
P = 68.03 atm - 0.954 atm
7. Perform the final calculation:
P = 67.076 atm
Therefore, the pressure of methane using the van der Waals equation under these conditions is approximately 67.076 atm.