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
Answers
Answered by
GPT 3.5
To calculate the pressure using the van der Waals equation, we first need to know the values of the van der Waals constants for neon gas:
a = 0.2136 L^2 atm/mol^2
b = 0.01718 L/mol
Now we can use the van der Waals equation:
(P + a(n/V)^2)(V - nb) = nRT
Where:
P = pressure
a = van der Waals constant
n = number of moles
V = volume
b = van der Waals constant
R = ideal gas constant (0.08206 L atm/mol K)
T = temperature in Kelvin
Given:
V = 22.4 L
n = 1 mol
R = 0.08206 L atm/mol K
T = 100 °C = 373 K
Substitute the values into the van der Waals equation:
(P + 0.2136(1/22.4)^2)(22.4 - 0.01718) = 1 * 0.08206 * 373
(P + 0.2136(0.04464)^2)(22.38282) = 30.42638
(P + 0.2136(0.002104)*22.38282) = 30.42638
(P + 0.2136(0.04710562)) = 30.42638
P + 0.01005 = 30.42638
P = 30.42638 - 0.01005
P = 30.41633 atm
Therefore, the pressure in the 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is approximately 30.42 atm.
a = 0.2136 L^2 atm/mol^2
b = 0.01718 L/mol
Now we can use the van der Waals equation:
(P + a(n/V)^2)(V - nb) = nRT
Where:
P = pressure
a = van der Waals constant
n = number of moles
V = volume
b = van der Waals constant
R = ideal gas constant (0.08206 L atm/mol K)
T = temperature in Kelvin
Given:
V = 22.4 L
n = 1 mol
R = 0.08206 L atm/mol K
T = 100 °C = 373 K
Substitute the values into the van der Waals equation:
(P + 0.2136(1/22.4)^2)(22.4 - 0.01718) = 1 * 0.08206 * 373
(P + 0.2136(0.04464)^2)(22.38282) = 30.42638
(P + 0.2136(0.002104)*22.38282) = 30.42638
(P + 0.2136(0.04710562)) = 30.42638
P + 0.01005 = 30.42638
P = 30.42638 - 0.01005
P = 30.41633 atm
Therefore, the pressure in the 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is approximately 30.42 atm.