(i)
Interpretation:
The viscosity of air at 273 K is to be calculated.
Concept introduction:
Viscosity is called the property of a fluid that is “a measure of its resistance to flow”. It is the consequence of a molecular attraction and denotes the internal friction of a moving fluid. A fluid with high viscosity resists motion due to its high molecular interaction. The substances with low intermolecular interaction have lower viscosity.
(i)
Answer to Problem 16A.7AE
The viscosity of air at 273 K is 12.67 μP_.
Explanation of Solution
The viscosity of air is calculated by the formula shown below.
η=(m3σ)(4RTπM)1/2 (1)
Where,
- R is the gas constant.
- T is the temperature.
- σ is the collision cross section area.
- M is the molar mass.
- m is equal to MNA.
The molar mass is 29.0 g mol−1=29.0×10−3 kg mol−1.
The value of m is calculated as shown below.
m=29.0×10−3 kg mol−16.022×1023 mol−1=29.0×1.6605×10−27 kg
The Avogadro’s number is 6.022×1023 mol−1.
The value of gas constant is 8.314 J/mol⋅K.
The value of σ is 0.40 nm2.
The conversion of nm2 into m2 is done as shown below.
1 nm=1×10−9 m1 nm2=(1×10−9)2 m21 nm2=1×10−18 m2
Therefore, the conversion of 0.40 nm2 into m2 is done as shown below.
0.40 nm2=0.40×10−18 m2
Also, 1 J=1 kg⋅m2/s2
Substitute the values of m, R, T, σ and M in equation (1) to calculate the viscosity of air.
η=(29.0×1.6605×10−27 kg3×0.40×10−18 m2)(4×8.314 J/mol⋅K×273 K3.14×29.0×10−3 kg mol−1)1/2=(4.81545×10−26 kg1.2×10−18 m2)(9078.888 kg⋅m2/s2/mol0.09106 kg mol−1)1/2=4.012×10−8×315.756 kg m−1 s−1=1.2670×10−5 kg m−1 s−1
The conversion of kg m−1 s−1 into μP is done as shown below.
1 μP=1×10−6 kg m−1 s−11 kg m−1 s−1=106 μP
Therefore, the conversion of 1.2670×10−5 kg m−1 s−1 into μP is done as shown below.
1.2670×10−5 kg m−1 s−1=1.2670×10−5×106 μP=12.67 μP_
Therefore, the viscosity of air at 273 K is 12.67 μP_.
(ii)
Interpretation:
The viscosity of air at 298 K is to be calculated.
Concept introduction:
As mentioned in the concept introduction in part (a).
(ii)
Answer to Problem 16A.7AE
The viscosity of air at 298 K is 13.235 μP_.
Explanation of Solution
The viscosity of air is calculated by the formula shown below.
η=(m3σ)(4RTπM)1/2 (1)
Where,
- R is the gas constant.
- T is the temperature.
- σ is the collision cross section area.
- M is the molar mass.
- m is equal to MNA.
The molar mass is 29.0 g mol−1=29.0×10−3 kg mol−1.
The value of m is calculated as shown below.
m=29.0×10−3 kg mol−16.022×1023 mol−1=29.0×1.6605×10−27 kg
The Avogadro’s number is 6.022×1023 mol−1.
The value of gas constant is 8.314 J/mol⋅K.
The value of σ is 0.40 nm2.
The conversion of nm2 into m2 is done as shown below.
1 nm=1×10−9 m1 nm2=(1×10−9)2 m21 nm2=1×10−18 m2
Therefore, the conversion of 0.40 nm2 into m2 is done as shown below.
0.40 nm2=0.40×10−18 m2
Also, 1 J=1 kg⋅m2/s2
Substitute the values of m, R, T, σ and M in equation (1) to calculate the viscosity of air.
η=(29.0×1.6605×10−27 kg3×0.40×10−18 m2)(4×8.314 J/mol⋅K×298 K3.14×29.0×10−3 kg mol−1)1/2=(4.81545×10−26 kg1.2×10−18 m2)(9910.288 kg⋅m2/s2/mol0.09106 kg mol−1)1/2=4.012×10−8×329.89 kg m−1 s−1=1.3235×10−5 kg m−1 s−1
The conversion of kg m−1 s−1 into μP is done as shown below.
1 μP=1×10−6 kg m−1 s−11 kg m−1 s−1=106 μP
Therefore, the conversion of 1.3235×10−5 kg m−1 s−1 into μP is done as shown below.
1.3235×10−5 kg m−1 s−1=1.3235×10−5×106 μP=13.235 μP_
Therefore, the viscosity of air at 298 K is 13.235 μP_.
(iii)
Interpretation:
The viscosity of air at 1000 K is to be calculated.
Concept introduction:
As mentioned in the concept introduction in part (a).
(iii)
Answer to Problem 16A.7AE
The viscosity of air at 1000 K is 24.245 μP_.
Explanation of Solution
The viscosity of air is calculated by the formula shown below.
η=(m3σ)(4RTπM)1/2 (1)
Where,
- R is the gas constant.
- T is the temperature.
- σ is the collision cross section area.
- M is the molar mass.
- m is equal to MNA.
The molar mass is 29.0 g mol−1=29.0×10−3 kg mol−1.
The value of m is calculated as shown below.
m=29.0×10−3 kg mol−16.022×1023 mol−1=29.0×1.6605×10−27 kg
The Avogadro’s number is 6.022×1023 mol−1.
The value of gas constant is 8.314 J/mol⋅K.
The value of σ is 0.40 nm2.
The conversion of nm2 into m2 is done as shown below.
1 nm=1×10−9 m1 nm2=(1×10−9)2 m21 nm2=1×10−18 m2
Therefore, the conversion of 0.40 nm2 into m2 is done as shown below.
0.40 nm2=0.40×10−18 m2
Also, 1 J=1 kg⋅m2/s2
Substitute the values of m, R, T, σ and M in equation (1) to calculate the viscosity of air.
η=(29.0×1.6605×10−27 kg3×0.40×10−18 m2)(4×8.314 J/mol⋅K×1000 K3.14×29.0×10−3 kg mol−1)1/2=(4.81545×10−26 kg1.2×10−18 m2)(33256 kg⋅m2/s2/mol0.09106 kg mol−1)1/2=4.012×10−8×604.325 kg m−1 s−1=2.4245×10−5 kg m−1 s−1
The conversion of kg m−1 s−1 into μP is done as shown below.
1 μP=1×10−6 kg m−1 s−11 kg m−1 s−1=106 μP
Therefore, the conversion of 2.4245×10−5 kg m−1 s−1 into μP is done as shown below.
2.4245×10−5 kg m−1 s−1=2.4245×10−5×106 μP=24.245 μP_
Therefore, the viscosity of air at 1000 K is 24.245 μP_.
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