As shown in the figure at $200 K$ and $500 a t m$ value of compressibility factor is $2$ (approx). Then the volume of the gas at this point will be:

0.01 L

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0.09 L

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0.065 L

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0.657 L

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Solution

The correct option is A $0.065 L$
$P V = Z R T \Rightarrow [For real Gases] (500) (V) = (2) (0.0821) (200) V = (\frac{400}{500}) (0.0821) = 0.065 l i t r e s$

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As shown in the figure at 200 K and 500 atm value of compressibility factor is 2(approx). Then the volume of the gas at this point will be:

The correct option is A 0.065 LPV=ZRT⇒[For real Gases](500)(V)=(2)(0.0821)(200)V=(400500)(0.0821)=0.065litres

As shown in the figure at $200 K$ and $500 a t m$ value of compressibility factor is $2$ (approx). Then the volume of the gas at this point will be:

The correct option is A $0.065 L$
$P V = Z R T \Rightarrow [For real Gases] (500) (V) = (2) (0.0821) (200) V = (\frac{400}{500}) (0.0821) = 0.065 l i t r e s$