Question

At higher pressure and at a temperature T = 273 K ,Z = (Compressibility factor). We have :-

$\frac{dz}{dp} = \frac{1}{2.8}$
What is the volume of one mole of gas.​

Answer 1

Answer:

2 L

Explanation:

We know,

Z = 1 + (Pb/RT)

Differentiating with respect to P

(dZ/dP) = (b/RT)

(1/2.8) = (b/RT) = dZ/dP

b = RT/2.8

b = 8

4 {Na * (4/3) * pi * r^3} = Volume of 1 L of gas

= 5.6/2.8 = 2 L

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Answer 2

Given

• At higher pressure, $\frac{dZ}{dP}\:=\:\frac{1}{2.8}$

To Find

Volume of one mole of gas.

Concept

• At lower pressure, we write

$\boxed{\green{Z\:=\:1\:-\:\frac{a}{RTV_m}}}$

While, at higher pressure, Z becomes

$\boxed{\green{Z\:=\:1\:+\:\frac{Pb}{RT}}}$

• a and b are Van der Waal's constant.

• a is a measure of force of attraction between gaseous molecule.More the value of a, more easily the gas is liquefiable

• b occurs in volume correction term. It is equal to four times the volume of one mole of gas.More the value of b ,less compressible is the gas.

Solutions

We have,

T = 273 K.

R = $\frac{22.4}{273}$ ( It will make the calculation easier)

At higher pressure region, $\boxed{\green{Z\:=\:1\:+\:\frac{Pb}{RT}}}$

Differentiating both side with respect to P (pressure)

=> $\frac{dZ}{dP} \:=\: \frac{b}{RT}$

=> $\frac{dZ}{dP} \:= \:\frac{b}{RT}$

=> $\frac{1}{2.8} \:= \:\frac{b}{ \frac{22.4}{273}\:×\:273 }$

=> $b\:=\: \frac{22.4}{2.8}$

=> $b\:=\: 8$

Hence, volume of 4 moles of gas = b = 8 litre

Thus, volume of 1 mole of gas = $\frac{8}{4}$ = 2 litre.

$\boxed{\boxed{\green{Volume\:of\:one\:mole\:of\:gas\:is\:2\:litres}}}$