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 $(300 K)$ and very high pressure, then calculate diameter of the molecules.
(Given: $N_{A} = 6.0 \times 10^{23}, R = 0.0821 a t m . l i t m o l^{- 1} K^{- 1}$ )

7.5 ˚ A

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5 ˚ A

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2.5 ˚ A

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1.25 ˚ A

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Solution

The correct option is B $5 ˚ A$
$Z = 1 + \frac{P b}{R T}$ , at high pressure. (always)
Slope $= \frac{b}{R T} = \frac{π}{492.6}$
$\Rightarrow b = \frac{π}{492.6} \times 0.0821 \times 300$ and
$b = \frac{4}{3} π r^{3} \times 4 N_{A}$
$= \frac{π}{492.6} \times 0.0821 \times 300 \times 10^{- 3}$
$\Rightarrow r = 2.5 ˚ A$ or $d = 5 ˚ A$ .

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Similar questions

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: $N_{A} = 6.023 \times 10^{23}, R = 0.0821 a t m L m o l^{- 1} K^{- 1}$ )

Q. f the slope of

Z

(compressibility factor) vs

P

curve be constant (slope

= π / 492.6 {a t m}^{- 1}

) at a particular temperature

300 K

and very high pressure. Calculate the diameters of molecules (given

N_{A} = 6 \times 10^{23}, R = 0.0821 L a t m . {m o l}^{- 1} K^{- 1}

Q. Which of the following options is correct about the plot between compressibility factor

Z

and

\frac{1}{T}

? (T= temperature in

K

)

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 a 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: $(i n L / m o l)$ $(G i v e n, N_{A} = 6 \times 10^{23}$ and $R = \frac{22.4}{273} L a t m K^{- 1} m o l^{- 1})$

State True or False.

In very high pressure region if

Z

v/s

P

is plotted at 1200 K for the above gas then it will have greatest slope.

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If the slope of ′Z′ (compressibility factor) v/s ′P′ curve is constant (slope =π492.6atm−1) at a particular temperature (300K) and very high pressure, then calculate diameter of the molecules.(Given: NA=6.0×1023,R=0.0821atm.litmol−1K−1)

The correct option is B 5˚AZ=1+PbRT, at high pressure. (always)Slope=bRT=π492.6⇒b=π492.6×0.0821×300 andb=43πr3×4NA=π492.6×0.0821×300×10−3⇒r=2.5˚A or d=5˚A.