The Van der Waals equation is written as: P n2a/V2 (V-nb) = nRT which relates th

Question

The Van der Waals equation is written as: P n2a/V2 (V-nb) = nRT which relates the pressure (P), volume (V) and Temperature (T) of a gas Here, n is the number of moles of gas and R is the universal gas constant (R = 0 08206 atm-liter (mole-K)). The term involving the parameter a corrects the pressure of intermolecular attractive forces, while the term involving the parameter b is a correction for that portion of the volume of the gas that is not compressible due to intrinsic volume of the gas molecules If given that one mole of Sarin gas has a pressure of 2 atm and a temperature of 313 K. and a = 629 atm-liter2/mole- and b = 0 0562 liter mole, find the volume of the Sarin gas, V. by using the Newton-Raphson method Use Vo, = X Do 3 iterations Find volume when X = 6 Volume after 3 iterations =

Explanation / Answer

Let's start by filling in all the numbers in this equation. A quick check shows that all the units match, so since we're concentrating on this as a math problem, I'm going to leave them off for now to avoid cluttering the equation.

R = .08206

P = 2

T = 313

n = 1 (specified to be 1 mole of Sarin gas)

a = 629

b = .0562

So P((n2a)/V2)(V-nb) = nRT becomes

2((12*629)/V2)(V-1*.0562) = 1*.08206*313

Multiply out to simplify:

(1258/V2)(V-.0562) = 25.68478

Distribute:

1258/V - 70.6996/V2 = 25.68478

Now multiply through by V2:

1258V - 70.6996 = 25.68478V2

or

0 = 25.68478V2 - 1258V + 70.6996

Now to find V.

Newton's method (aka the Newton-Raphson Method) allows us to approximate the zeroes of a function. We start with a guess, and then adjust that guess through each iteration by the following formula:

xn+1 = xn - f(xn)/f'(xn)

Geometrically, you can think of this as looking at the xn we have, finding the corresponding point on the graph, drawing a tangent line at that point, finding where that tangent line intersects the x-axis, and using that as our next guess for the zero and assigning it as xn+1.

For this equation, f(x) = 25.68478V2 - 1258V + 70.6996, as solving for the zeroes of that will give us solutions to the original equation. Taking the derivative, f'(x) = 51.36956V - 1258, and the problem gives us a first guess for a zero of V0 = 6.

First iteration:

V1 = V0 - f(V0)/f'(V0)

V1 = 6 - f(6)/f'(6)

V1 = 6 - (25.68478(6)2 - 1258(6) + 70.6996)/(51.36956(6) - 1258)

V1 = -.89910306

Second iteration:

V2 = V1 - f(V1)/f'(V1)

V2 = -.89910306 - f(-.89910306)/f'(-.89910306)

V2 = -.89910306 - (25.68478(-.89910306)2 - 1258(-.89910306) + 70.6996)/(51.36956(-.89910306) - 1258)

V2 = .03828929

Third iteration:

V3 = V2 - f(V2)/f'(V2)

V3 = .03828929 - f(.03828929)/f'(.03828929)

V3 = .03828929 - (25.68478(.03828929)2 - 1258(.03828929) + 70.6996)/(51.36956(.03828929) - 1258)

V3 = .05625803

So the value for V after three iterations is .05625803 liters.

(Note: A quick solve with the quadratic formula gives .05626463 as one of the solutions -- so the third iteration is already very close to the actual root!)

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