Decomposition reaction: CH2=CH2(g)---> CH4(g) + C (Graphite)

If the decomposition begins at 10c and 40.0 atm with a gas density of .215 g/ml and the temp increases by 950K. (a) what is the final temperature of the confined gas (ignore volume of graphite and use the van der waals equation)?

I have this for van der waals: (P + (n^2a/V^2)) * (V -nb) = nRT

I know my a and b constants. a = 2.25, b = .0428.

Do I need to use the density to find n? .215g/ml = 215g/L. Can I assume that there is 1 Liter in my equation. If I do that then there are 13.4 moles in the gas.

This would make my equation (P + (3.14^2*2.25/1^2)*(1-3.14*.0428)= 3.14*.0821*1233

This equation doesnt seem right. Can anyone help me figure out where I'm making my mistakes?

anyone?

To find the final temperature of the confined gas using the van der Waals equation, you'll need to follow a few steps:

Step 1: Calculate the initial number of moles (n) of the gas using the given density. Since the density is 0.215 g/ml, you need to convert it to g/L. Since 1 ml = 1 cm^3 = 1/1000 L, the density is equivalent to 215 g/L. Now, divide the mass of the gas by its molar mass to get the number of moles. Let's assume the molar mass of CH2=CH2(g) is 28 g/mol (you can verify this).

n = mass/molar mass
n = 215 g/ 28 g/mol
n ≈ 7.68 mol

Step 2: Substitute the values into the van der Waals equation.

(P + (n^2a/V^2)) * (V - nb) = nRT

Let's assume the initial pressure (P) is 40.0 atm, the initial volume (V) is 1 L, and the constants a = 2.25 and b = 0.0428.

(40.0 + (7.68^2 * 2.25/1^2)) * (1 - 7.68 * 0.0428) = 7.68 * R * T

Simplifying this equation will give you:

(40 + 132.9504) * (1 - 0.327264) = 7.68 * R * T
(172.9504) * (0.672736) = 7.68 * R * T
116.287265144 = 7.68 * R * T

Step 3: Rearrange the equation to solve for the final temperature (T).

T = 116.287265144 / (7.68 * R)

Assuming R is the ideal gas constant, R = 0.0821 L*atm/(mol*K), we can substitute it into the equation:

T = 116.287265144 / (7.68 * 0.0821)
T ≈ 1825 K

Therefore, the final temperature of the confined gas is approximately 1825 K.