
At what pressure and $ {127^ \circ }C $ , the density of $ {O_2} $ gas becomes $ 1.6g/L $ ? [ $ a = 4.0atm{L^2}mo{l^{ - 2}},b = 0.4Lmo{l^{ - 1}},R = 0.08atm{K^{ - 1}}mo{l^{ - 1}} $ ]
Answer
497.7k+ views
Hint: 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 = 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: $ \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^ \circ }C = 127 + 273 = 400K $ ,
$ Density(d) = 1.6g/L = \dfrac{M}{V} $ $ V = \dfrac{M}{d} = \dfrac{{32}}{{1.6}}L $ (Molecular weight of Oxygen is 32g/mol)
$ a = 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,
$ \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 = 32 - 19.6 = 1.6326 $
$ P = 1.6227atm $
The pressure at which the density of $ {O_2} $ gas becomes $ 1.6g/L $ is $ 1.6227atm $
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.
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 = 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: $ \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^ \circ }C = 127 + 273 = 400K $ ,
$ Density(d) = 1.6g/L = \dfrac{M}{V} $ $ V = \dfrac{M}{d} = \dfrac{{32}}{{1.6}}L $ (Molecular weight of Oxygen is 32g/mol)
$ a = 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,
$ \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 = 32 - 19.6 = 1.6326 $
$ P = 1.6227atm $
The pressure at which the density of $ {O_2} $ gas becomes $ 1.6g/L $ is $ 1.6227atm $
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

