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  • If density of vapours of a substance of Molar mass 18 gm mole at 1 atm pressure and 500 K is <img alt="0.36 \mathrm{~kg} \mathrm{~m}^{-3}" src="https://entrancecorner.oncodecogs.com/gif.l

If density of vapours of a substance of Molar mass 18 gm mole at 1 atm pressure and 500 K is 0.36 \mathrm{~kg} \mathrm{~m}^{-3} then value of Z for the vapour is?

Option: 1

\frac{50}{41}


Option: 2

1.8


Option: 3

9


Option: 4

\frac{41}{50}


Answers (1)

\mathrm{\text { Vreal }=\frac{\text { Molar Mass }}{\text { density}}=\frac{18}{.36}}

\mathrm{\text { Videal }=\frac{n R T}{P}=\frac{1 \times0.082 \times 500}{1}}

\mathrm{\text { so, } z=\frac{v_{\text {real }}}{v_{\text {ideal }}}=\frac{50}{.082 \times 500}=\frac{50}{41}}

Posted by

Ramraj Saini

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