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  • A gas ras a compressibility factor ot 1 and a molor volume at <img alt="2 \mathrm{dm}^{3} \mathrm{~mol}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20%5Cmathrm%7Bdm%7D%5E%7B3%7D%20

A gas ras a compressibility factor ot 1 and a molor volume at 2 \mathrm{dm}^{3} \mathrm{~mol}^{-1} at a Temperature of 800 \mathrm{k} and pressure A atm
If it shows ideal gas behavaiar at the same teperature and pressure
The molar volume will be \mathrm{B\, dm^{3}mol^{-}}
the value at \mathrm{\frac{A}{B}\, is-}

[use gas constant \mathrm{=8 \times 10^{-2} \mathrm{Latm}\, \mathrm{K}^{-1} \text { mol }^{-1}} ]

Option: 1

1600


Option: 2

3200


Option: 3

800


Option: 4

80


Answers (1)

best_answer

\text{compressibility factor }(z)=\frac{\text { Vreal }}{\text { videal }}=1

\text { Vreal }= 2 \mathrm{dm}^{3} \mathrm{Mol}^{-1}= 2 \mathrm{~L} \text { /mol }
\text { videal }=\frac{2}{1}=2 \, L/ \mathrm{mol}
\therefore B=2 L / \mathrm{~mol}

Using ideal gas equation \mathrm{P V=n R T}
                                         \mathrm{P=\frac{1 \times 8 \times 10^{-2} \times 800}{2}}
                                          \mathrm{P=32 \times 10^{1}=320 \mathrm{Atm} }

\mathrm{\frac{A}{B}=\frac{320}{2}=1600 }

\mathrm{\text { Volue of } \frac{A}{B}=1600 }

Posted by

Kuldeep Maurya

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