Question

Use the van der Waals equation to calculate the temperature of 1.302 moles of chlorine gas contained in a 4.725-liter flask at 8.7165 atmospheres. Given: a = 6.49 L2 · atm/mol2, b = 0.0562 L/mol, and R = 0.08205 L · atm/K · mol A. 2.847 K B. 3.569 K C. 401.1 K D. 385.52 K

Answer 1

B. 356.9 K

Step-by-step explanation:

Hello,

In this case, considering the Van der Waals equation shown below:

`P=(RT)/(v-b)-(a)/(v^2)`

Solving for T one obtains:

`T=((v-b))/(R)(P-(a)/(v^2) )`

The molar volume is:

`v=(4.725L)/(1.302mol)=3.63L/mol`

So the temperature turns out:

`T=(3.63mol/L-0.0562mol/L)/(0.082 (atm*L)/(mol*K))*(8.7165atm-(6.49L^2*atm/mol^2)/((3.63mol/L)^2) )\\T=358.42K`

Therefore the answer should be B which is closer to the obtained result.

Best regards.