If at 200 K and 500 $atm$ , density of ${CH}_{4}$ is 0.246 $g / ml$ , then its compressibility factor {Z) is approximately $2.0 \times 10^{x}$ then $x$ is:

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Detailed Solution

The density can be given as;

$ρ = \frac{P M}{Z R T}$

$\begin{array}{l} Z = \frac{P M}{ρ R T} = \frac{\frac{500 \times 16}{0.246 \times 10^{+ 3}}}{0.0821 \times 200} = 2 \times 10^{0} \\ x = 0 \end{array}$

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