Answer:
B. 356.9 K
Explanation:
Hello,
In this case, considering the Van der Waals equation shown below:
[tex]P=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]
Solving for T one obtains:
[tex]T=\frac{(v-b)}{R}(P-\frac{a}{v^2} )[/tex]
The molar volume is:
[tex]v=\frac{4.725L}{1.302mol}=3.63L/mol[/tex]
So the temperature turns out:
[tex]T=\frac{3.63mol/L-0.0562mol/L}{0.082 \frac{atm*L}{mol*K}}*(8.7165atm-\frac{6.49L^2*atm/mol^2}{(3.63mol/L)^2} )\\T=358.42K[/tex]
Therefore the answer should be B which is closer to the obtained result.
Best regards.
