Question

The compressibility factor for $N_2$ at $223K$ and $81.06MPa$ is $1.95$, and at $373K$ and $20.265$ MPa, it is $1.10$. A certain mass of $N_2$ occupies a volume of $1.0d{m}^3$ at $223K$ and $81.06$ MPa. What is the volume occupied by the same quantity of $N_2$ at $373K$ and $20.265$ MPa?

• A. $3.774d{m}^3$
• B. $2.77d{m}^3$
• C. $5.07d{m}^3$
• D. $9.30d{m}^3$

Answer 1

Answer: (A)

For $T=223K$, $P=81.06$ Mpa, $Z=1.95$, and $V=1.0d{m}^3$ $=10^{3}c{m}^3$, we have
$n=\frac{PV}{ZRT}$ $=\frac{81.06\times 10^3}{1.95\times 8.314\times 223}=22.42$ mole
Now, at $T=373K$, $P=20.265$ MPa, $Z=1.10$, the volume occupied will be
$V=\frac{ZnRT}{P}$ $=\frac{1.10\times 22.42\times 8.314\times 373}{20.265}$ $=3774.0c{m}^3$
$\therefore$ $V=3.774$ $d{m}^3$