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Standard XII

Physics

Expression for Standing Waves

The graph of ...

Question

The graph of compressibility factor $(Z) v s P$ for one mole of a real gas is shown in following diagram. The graph is plotted at constant temperature $273 K$ . If the slope of graph at very high pressure $(\frac{d Z}{d P})$ is $(\frac{1}{2.8}) a t m^{- 1}$ , then the volume of one mole of real gas molecules is: (in $L / m o l$ ) [Given, $N_{A} = 6 \times 10^{23}$ and $R = \frac{22.4}{273} L a t m K^{- 1} m o l^{- 1}$ ]

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Solution

At very high pressure, $(\frac{d Z}{d P}) = (\frac{1}{2.8}) a t m^{- 1}; V / R T = 1 / 2.8, V = 8$ .

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Similar questions

Q. The graph of compressibility factor

(Z) v s P

for one mole of a real gas is shown in following diagram. The graph is plotted at constant temperature

273 K

. If the slope of graph at very high pressure

(\frac{d Z}{d P})

(\frac{1}{2.8}) a t m^{- 1}

, then the volume of one mole of real gas molecules is: (in

L / m o l

) [Given,

N_{A} = 6 \times 10^{23}

and

R = \frac{22.4}{273} L a t m K^{- 1} m o l^{- 1}

]

Q. f the slope of

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= π / 492.6 {a t m}^{- 1}

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If the slope of ‘Z’ (compressibility factor ) v/s ‘P’ curve is constant (slope = $\frac{π}{492.6} a t m^{- 1}$ ) at a particular temperature (300K) and very high pressure. Calculate diameter of the molecules. (Given: $N_{A} = 6.023 \times 10^{23}, R = 0.0821 a t m L m o l^{- 1} K^{- 1}$ )

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At very high pressure, (dZdP)=(12.8)atm−1;V/RT=1/2.8,V=8.

At very high pressure, $(\frac{d Z}{d P}) = (\frac{1}{2.8}) a t m^{- 1}; V / R T = 1 / 2.8, V = 8$ .