15.0 moles of gas are in a 8.00 L tank at 21.7 ∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol .
What is your problem with this? Just substitute the number and go through the math. Surely you don't need someone to do the math for you. It's arithmetic.
To calculate the difference in pressure between methane and an ideal gas under these conditions, we can use the van der Waals equation:
(P + n^2a/V^2)(V - nb) = nRT
Where:
P = pressure
n = number of moles of the gas
V = volume
a, b = van der Waals constants
R = ideal gas constant
T = temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(Kelvin) = T(Celsius) + 273.15
So, T = 21.7 + 273.15 = 294.85 K
Given:
n = 15.0 moles
V = 8.00 L
a = 2.300 L^2.atm/mol^2
b = 0.0430 L/mol
R = 0.0821 L.atm/(mol.K)
Now, let's substitute the values into the van der Waals equation:
(P + (n^2a / V^2))(V - nb) = nRT
Expanding and rearranging the equation, we get:
PV - Pnb + (n^2a / V) - (n^2a * b / V^2) = nRT
Simplifying further:
PV - Pnb + (n^2a / V) - (n^2a * b / V^2) = nRT
We need to solve for P, the pressure difference between methane and an ideal gas. To do this, we can rearrange the equation as follows:
P = (nRT + n^2a / V^2)(V - nb) / V
Now, let's substitute the known values:
P = (15.0 * 0.0821 * 294.85 + (15.0^2 * 2.300) / (8.00^2))(8.00 - 15.0 * 0.0430) / 8.00
Evaluating this expression will yield the difference in pressure between methane and an ideal gas under these conditions.
To calculate the difference in pressure between methane (an real gas) and an ideal gas under these conditions, we can use the Van der Waals equation.
The Van der Waals equation is given by:
(P + a(n^2/V^2))(V - nb) = nRT
Where:
P = pressure
n = number of moles of gas
V = volume of the gas
a = Van der Waals constant related to attractive forces between gas molecules
b = Van der Waals constant related to the volume of one mole of gas molecules
R = ideal gas constant
T = temperature
Given data:
n = 15.0 moles
V = 8.00 L
T = 21.7 ∘C = (21.7 + 273.15) K (Convert to Kelvin)
a = 2.300 L^2⋅atm/mol^2
b = 0.0430 L/mol
R = 0.0821 L⋅atm/mol⋅K (Ideal gas constant)
Let's plug in the values into the equation and solve for the pressure difference:
First, convert the temperature to Kelvin:
T = 21.7 + 273.15 = 294.85 K
(Van der Waals equation) * (V - nb) = nRT
(P + a(n^2/V^2))(V - nb) = nRT
P(V - nb) + a(n^2/V) = nRT
PV - Pnb + an^2/V = nRT
PV = Pnb - an^2/V + nRT
PV = (Pnb + nRT) - an^2/V
PV = (Pnb + nRT)/V - an^2/V
PV = (Pn + n^2/V^2)b - an^2/V
P = (n/V)(RT/1 - n/Vb) - an^2/V
Now we can calculate the pressure for methane using this equation:
P_methane = (n/V)(RT/1 - n/Vb) - an^2/V
P_methane = (15.0 moles / 8.00 L)(0.0821 L⋅atm/mol⋅K)(294.85 K / 1 - (15.0 moles / 8.00 L)(0.0430 L/mol)) - (2.300 L^2⋅atm/mol^2)(15.0 moles^2 / 8.00 L)
P_methane = 4.522 atm
Now, let's calculate the pressure for an ideal gas at the same conditions:
Ideal Gas Law equation:
PV = nRT
P_ideal = (n/V)(RT)
P_ideal = (15.0 moles / 8.00 L)(0.0821 L⋅atm/mol⋅K)(294.85 K)
P_ideal = 5.125 atm
Finally, we can find the difference in pressure between methane and an ideal gas:
Difference in pressure = P_methane - P_ideal
Difference in pressure = 4.522 atm - 5.125 atm
Difference in pressure = -0.603 atm
Therefore, the difference in pressure between methane and an ideal gas under these conditions is approximately -0.603 atm.