Question

$C{O}_2$ gas is liquefied at $27^{o}C$ . Gas follows the following graph during liquefaction for 1 mole of $C{O}_2$ .



Then which statement is (are) correct in gaseous phase will

(i) The maximum density of gas is 0.1 gm/ml

(ii) The density of liquid $C{O}_2$ is 1 g/ml at 60 atm.

(iii) At point C 50% of $C{O}_2$ is liquefied.

(iv) The compressibility factor of gas at $27^{o}C$ is always less than 1.

Which of the above is/are correct

• A. only i
• B. i and ii
• C. i,ii and iii
• D. all

Answer 1

Answer: (C)

$\left(A\right)$ Maximum density of $C{O}_2$ in gaseous phase will be at point 'A'

Now: $d=\frac{PM}{RT}=\frac{60\times 44}{R\times 300}=107gm/lt=0.107gm/cc$

Or $d=\frac{massofgas}{Volumeofgas}=\frac{44gm}{440cc}=0.1$

$\left(B\right)$ Density of liquid $C{O}_2$ at 60 atm (at point B)

At point B there will be only liquid $C{O}_2$ . So we can find out the density of density of liquid $C{O}_2$ only at point B.

$\text{d}=\frac{\text{M}}{\text{V}}=\frac{44}{0.044}=1\text{ gm/cc}$

$\left(C\right)$ at point C the volume will be $\frac{0.44+0.044}{2}\approx 0.22\text{L}$

And pressure is same so obviously approx. half of $C{O}_2$ is liquified

$\left(D\right)$ Z can be greater than 1 at $27^{o}C$ . For example at point 'A' Z is 1.07, So wrong.