Solveeit Logo
Login

Question

Question: At what pressure and \( {127^ \circ }C \) , the density of \( {O_2} \) gas becomes \( 1.6g/L \) ? [ ...

At what pressure and 127∘C{127^ \circ }C , the density of O2{O_2} gas becomes 1.6g/L1.6g/L ? [ a=4.0atmL2mol−2,b=0.4Lmol−1,R=0.08atmK−1mol−1a = 4.0atm{L^2}mo{l^{ - 2}},b = 0.4Lmo{l^{ - 1}},R = 0.08atm{K^{ - 1}}mo{l^{ - 1}} ]

Explanation

Solution

This question is based on the Van der Waals equation for Real gases. The Van der Waals equation is based on the theory for fluids and liquids assuming the particles to occupy non-zero volume and subject to an interparticle attractive force.

Complete Step By Step Answer:
The Van der Waals Equation for Real gases is a derivative of the equation for the Ideal Gases. Let us recall the Ideal Gas equation: PV=nRTPV = nRT
Where, P is the pressure, V is the volume, T is the temperature, n is the no. of moles and R is the gas constant.
The Van Der Waals Equation for Real Gases considers the volume occupied by the particles. Hence the total volume becomes less than the volume of the container. This is known as Volume correction denoted by ‘b’. Similarly, it has a pressure correction denoted by ‘a’.
The modified equation now becomes: (P−an2V2)(V−nb)=nRT\left( {P - a\dfrac{{{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT (where a and b are correction constants)
The information given to us is:
T=127∘C=127+273=400KT = {127^ \circ }C = 127 + 273 = 400K ,
Density(d)=1.6g/L=MVDensity(d) = 1.6g/L = \dfrac{M}{V} V=Md=321.6LV = \dfrac{M}{d} = \dfrac{{32}}{{1.6}}L (Molecular weight of Oxygen is 32g/mol)
a=4.0atmL2mol−2,b=0.4Lmol−1,R=0.08atmK−1mol−1a = 4.0atm{L^2}mo{l^{ - 2}},b = 0.4Lmo{l^{ - 1}},R = 0.08atm{K^{ - 1}}mo{l^{ - 1}}
The no. of moles isn’t given, hence consider it as 1 mole.
Substituting the values in the Real Gas equation, and finding the value for Pressure P,
(P+4×1(321.6)2)(321.6−(1)(0.4))=1×0.08×400\left( {P + 4 \times \dfrac{1}{{{{\left( {\dfrac{{32}}{{1.6}}} \right)}^2}}}} \right)\left( {\dfrac{{32}}{{1.6}} - (1)(0.4)} \right) = 1 \times 0.08 \times 400
[P+0.01](19.6)=32[P + 0.01](19.6) = 32
P+0.01=32−19.6=1.6326P + 0.01 = 32 - 19.6 = 1.6326
P=1.6227atmP = 1.6227atm
The pressure at which the density of O2{O_2} gas becomes 1.6g/L1.6g/L is 1.6227atm1.6227atm

Note:
The correction constant a and b always have positive values and are characteristic of individual gases. If the correction factors are negligible then all Real Gases can be considered as ideal gases; Although all real gases behave as ideal gases at High Temperatures and Low pressures.