Question
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. ( a
= 0.211 L2-atm/mol2, b
= 0.0171 L/mol)
= 0.211 L2-atm/mol2, b
= 0.0171 L/mol)
Answers
Answered by
GPT 3.5
To calculate the pressure using the van der Waals equation, we first need to rearrange the equation to solve for pressure. The van der Waals equation is:
(P + a(n/V)^2) (V - nb) = nRT
Where:
P is the pressure
a and b are van der Waals constants
n is the number of moles
V is the volume
R is the ideal gas constant
T is the temperature
Plugging in the values given:
a = 0.211 L^2-atm/mol^2
b = 0.0171 L/mol
n = 1.00 mol
V = 22.4 L
R = 0.0821 L-atm/mol-K
T = 100°C = 373 K
Now we can substitute the values into the van der Waals equation and solve for pressure:
(P + (0.211 (1.00/22.4)^2)) (22.4 - 0.0171(1.00)) = 1.00 * 0.0821 * 373
(P + 0.00437)(22.3829) = 30.6233
P + 0.0973 = 30.6233
P = 30.6233 - 0.0973
P = 30.526 atm
Therefore, the pressure in the vessel containing 1.00 mol of neon gas at 100 °C is 30.526 atm.
(P + a(n/V)^2) (V - nb) = nRT
Where:
P is the pressure
a and b are van der Waals constants
n is the number of moles
V is the volume
R is the ideal gas constant
T is the temperature
Plugging in the values given:
a = 0.211 L^2-atm/mol^2
b = 0.0171 L/mol
n = 1.00 mol
V = 22.4 L
R = 0.0821 L-atm/mol-K
T = 100°C = 373 K
Now we can substitute the values into the van der Waals equation and solve for pressure:
(P + (0.211 (1.00/22.4)^2)) (22.4 - 0.0171(1.00)) = 1.00 * 0.0821 * 373
(P + 0.00437)(22.3829) = 30.6233
P + 0.0973 = 30.6233
P = 30.6233 - 0.0973
P = 30.526 atm
Therefore, the pressure in the vessel containing 1.00 mol of neon gas at 100 °C is 30.526 atm.