Question

At moderate pressure, the compressibility factor for a particular gas is given by: $\text{Z = 1 + 0}\text{.3p -}\frac{160p}{T}$ ($p$ in bar and $T$ in kelvin). what is the Boyle's temperature of this gas?

• A. $533K$
• B. $340K$
• C. $470K$
• D. $680K$

Answer 1

The correct option is A $533K$

Solution:-

At Boyle's temperature, the gas behaves as an ideal gas, i.e., the compressibility factor will be $1$.

Therefore,

$Z=1+0.3p-\frac{160p}{T}\left(\text{Given}\right)$

$\Rightarrow 0.3p-\frac{160p}{T}=0(\because Z=1)$

$\therefore 0.3=\frac{160}{T}$

$\Rightarrow T=\frac{160}{0.3}=533.33K$