(i)
Interpretation:
The diffusion constant of argon at 20 °C and 1.00 Pa is to be calculated. The flow of gas due to diffusion at 1.00 Pa has to be calculated.
Concept introduction:
Diffusion of gas is a process in which the gas follows from higher pressure area to lower pressure area. The process of diffusion requires a potential gradient. The diffusion coefficient of a gas is given by the expression as shown below.
D=13(kTσp)(8RTπM)1/2
(i)
Answer to Problem 16A.2AE
The diffusion constant of argon at 20 °C and 1.00 Pa is 147.69m2s−1_. The flow of gas due to diffusion at 1.00 Pa is −6062.8molm−2s−1_.
Explanation of Solution
The diffusion coefficient of a gas is given by the expression as shown below.
D=13(kTσp)(8RTπM)1/2 (1)
Where,
- R is the gas constant (8.314 J K−1 mol−1).
- T is the temperature.
- k is the Boltzmann constant (1.381×10−23 J K−1).
- σ is the collision cross section area.
- M is the molar mass.
- p is the pressure.
The temperature of argon gas is 20 °C.
The conversion of temperature in Kelvin is shown below.
T=(273+20.0 °C) K=293 K
The pressure of argon gas at 20 °C is 1.00 Pa.
The conversion of pressure in kg m−1 s−2 is shown below.
p=(1.00 Pa)(1 kg m−1 s−21 Pa)=1.00 kg m−1 s−2
The value of σ for argon has is 0.36 nm2.
The conversion of σ in m2 is shown below.
σ=(0.36 nm2)(1 m21020 nm)=0.36×10−20 m2
The molar mass of argon is 39.948×10−3 kg mol−1.
Substitute the value of R, T, k, σ, M, and p in the equation (1).
D=(13((1.381×10−23 J K−1)(293 K)(0.36×10−20 m2)(1.00 kg m−1 s−2))(8(8.314 J K−1 mol−1)(1 kg m2 s−21 J)(293 K)(3.14)(39.948×10−3 kg mol−1))1/2)=(0.3747)(155361.3328)1/2 m2 s−1=(0.3747)(394.159) m2 s−1=147.69m2s−1_
Therefore, the diffusion constant of argon at 20 °C and 1.00 Pa is 147.69m2s−1_.
The pressure gradient of the pipe is 1.0 bar m−1.
The flow of gas due to diffusion (JzNA) is given by the expression as shown below.
JzNA=−(DRT)dpdz
Where,
- dpdz is the pressure gradient.
Substitute the values of dpdz, D, R, and T in the above equation.
JzNA=−(147.69 m2 s−1(8.314 J K−1 mol−1)(1 kg m2 s−21 J)(293 K))(1.0 bar m−1)=−(0.0606 mol s kg−1)(1.0 bar m−1)(105 kg m−1 s−21 bar)=−6062.8molm−2s−1_
Therefore, the flow of gas due to diffusion at 1.00 Pa is −6062.8molm−2s−1_.
(b)
Interpretation:
The diffusion constant of argon at 20 °C and 100 kPa is to be calculated. The flow of gas due to diffusion at 100 kPa has to be calculated.
Concept introduction:
As mentioned in the concept introduction in part (a).
(b)
Answer to Problem 16A.2AE
The diffusion constant of argon at 20 °C and 100 kPa is 1.4769×10−3m2s−1_. The flow of gas due to diffusion at 100 kPa is −6.06289×10-7molm−2s−1_.
Explanation of Solution
The diffusion coefficient of a gas is given by the expression as shown below.
D=13(kTσp)(8RTπM)1/2 (1)
Where,
- R is the gas constant (8.314 J K−1 mol−1).
- T is the temperature.
- k is the Boltzmann constant (1.381×10−23 J K−1).
- σ is the collision cross section area.
- M is the molar mass.
- p is the pressure.
The temperature of argon gas is 20 °C.
The conversion of temperature in Kelvin is shown below.
T=(273+20.0 °C) K=293 K
The pressure of argon gas at 20 °C is 100 kPa.
The conversion of pressure in kg m−1 s−2 is shown below.
p=(100 kPa)(103 kg m−1 s−21 kPa)=1.00×105 kg m−1 s−2
The value of σ for argon has is 0.36 nm2.
The conversion of σ in m2 is shown below.
σ=(0.36 nm2)(1 m21020 nm)=0.36×10−20 m2
The molar mass of argon is 39.948×10−3 kg mol−1.
Substitute the value of R, T, k, σ, M, and p in the equation (1).
D=(13((1.381×10−23 J K−1)(293 K)(0.36×10−20 m2)(1.00×105 kg m−1 s−2))(8(8.314 J K−1 mol−1)(1 kg m2 s−21 J)(293 K)(3.14)(39.948×10−3 kg mol−1))1/2)=(3.747×10−6)(155361.3328)1/2 m2 s−1=(3.747×10−6)(394.159) m2 s−1=1.4769×10−3m2s−1_
Therefore, the diffusion constant of argon at 20 °C and 100 kPa is 1.4769×10−3m2s−1_.
The pressure gradient of the pipe is 1.0 bar m−1.
The flow of gas due to diffusion (JzNA) is given by the expression as shown below.
JzNA=−(DRT)dpdz
Where,
- dpdz is the pressure gradient.
Substitute the values of dpdz, D, R, and T in the above equation.
JzNA=−(1.4769×10−3 m2 s−1(8.314 J K−1 mol−1)(1 kg m2 s−21 J)(293 K))(1.0 bar m−1)=−(6.062 mol s kg−1)(1.0 bar m−1)(105 kg m−1 s−21 bar)=−6.06289×10-7molm−2s−1_
Therefore, the flow of gas due to diffusion at 100 kPa is −6.06289×10-7molm−2s−1_.
(c)
Interpretation:
The diffusion constant of argon at 20 °C and 10.0 MPa is to be calculated. The flow of gas due to diffusion at 10.0 MPa has to be calculated.
Concept introduction:
As mentioned in the concept introduction in part (a).
(c)
Answer to Problem 16A.2AE
The diffusion constant of argon at 20 °C and 10.0 MPa is 1.4769×10−5m2s−1_. The flow of gas due to diffusion at 10.0 MPa is −6.0628×10−9molm−2s−1_.
Explanation of Solution
The diffusion coefficient of a gas is given by the expression as shown below.
D=13(kTσp)(8RTπM)1/2 (1)
Where,
- R is the gas constant (8.314 J K−1 mol−1).
- T is the temperature.
- k is the Boltzmann constant (1.381×10−23 J K−1).
- σ is the collision cross section area.
- M is the molar mass.
- p is the pressure.
The temperature of argon gas is 20 °C.
The conversion of temperature in Kelvin is shown below.
T=(273+20.0 °C) K=293 K
The pressure of argon gas at 20 °C is 10.0 MPa.
The conversion of pressure in kg m−1 s−2 is shown below.
p=(10.0 MPa)(106 kg m−1 s−21 MPa)=1.00×107 kg m−1 s−2
The value of σ for argon has is 0.36 nm2.
The conversion of σ in m2 is shown below.
σ=(0.36 nm2)(1 m21020 nm)=0.36×10−20 m2
The molar mass of argon is 39.948×10−3 kg mol−1.
Substitute the value of R, T, k, σ, M, and p in the equation (1).
D=(13((1.381×10−23 J K−1)(293 K)(0.36×10−20 m2)(1.00×107 kg m−1 s−2))(8(8.314 J K−1 mol−1)(1 kg m2 s−21 J)(293 K)(3.14)(39.948×10−3 kg mol−1))1/2)=(3.747×10−8)(155361.3328)1/2 m2 s−1=(3.747×10−8)(394.159) m2 s−1=1.4769×10−5m2s−1_
Therefore, the diffusion constant of argon at 20 °C and 10.0 MPa is 1.4769×10−5m2s−1_.
The pressure gradient of the pipe is 1.0 bar m−1.
The flow of gas due to diffusion (JzNA) is given by the expression as shown below.
JzNA=−(DRT)dpdz
Where,
- dpdz is the pressure gradient.
Substitute the values of dpdz, D, R, and T in the above equation.
JzNA=−(1.4769×10−5 m2 s−1(8.314 J K−1 mol−1)(1 kg m2 s−21 J)(293 K))(1.0 bar m−1)=−(0.0606 mol s kg−1)(1.0 bar m−1)(105 kg m−1 s−21 bar)=−6.0628×10−9molm−2s−1_
Therefore, the flow of gas due to diffusion at 10.0 MPa is −6.0628×10−9molm−2s−1_.
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Chapter 16 Solutions
Atkins' Physical Chemistry
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