Question

At moderate pressure, the compressibility factor for a particular gas is given by:
$z=1+0.34p-\frac{160p}{T}$ (p in bar and T in kelvin)
What is the Boyle's temperature of this gas?

• A. 298 K
• B. 340 K
• C. 470 K
• D. 680 K

Answer 1

Answer: (A), (C)

The correct options are
A 298 K
C 470 K

Given:

$z=1+0.34p-\frac{160p}{T}$

At Boyle’s temperature real gas behaves as Ideal gas.

Therefore $z=1$

$\Rightarrow 1=1+0.34p-\frac{160p}{T}$

$\Rightarrow 0=0.34p-\frac{160p}{T}$

$\Rightarrow \frac{160}{T}=0.34$

$\Rightarrow T=\frac{160}{0.34}$

$T=470K$

The correct answer is option C.