The diameter of bubble as its reaches the surface.

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Answer 1

Question

Definition Definition Law that is the combined form of Boyle's Law, Charles's Law, and Avogadro's Law. This law is obeyed by all ideal gas. Boyle's Law states that pressure is inversely proportional to volume. Charles's Law states that volume is in direct relation to temperature. Avogadro's Law shows that volume is in direct relation to the number of moles in the gas. The mathematical equation for the ideal gas law equation has been formulated by taking all the equations into account: PV=nRT Where P = pressure of the ideal gas V = volume of the ideal gas n = amount of ideal gas measured in moles R = universal gas constant and its value is 8.314 J.K-1mol-1 T = temperature

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm

Therefore, the diameter is 2.1 mm.

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Answer 2

Question

Definition Definition Law that is the combined form of Boyle's Law, Charles's Law, and Avogadro's Law. This law is obeyed by all ideal gas. Boyle's Law states that pressure is inversely proportional to volume. Charles's Law states that volume is in direct relation to temperature. Avogadro's Law shows that volume is in direct relation to the number of moles in the gas. The mathematical equation for the ideal gas law equation has been formulated by taking all the equations into account: PV=nRT Where P = pressure of the ideal gas V = volume of the ideal gas n = amount of ideal gas measured in moles R = universal gas constant and its value is 8.314 J.K-1mol-1 T = temperature

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm

Therefore, the diameter is 2.1 mm.

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Answer 3

Question

Definition Definition Law that is the combined form of Boyle's Law, Charles's Law, and Avogadro's Law. This law is obeyed by all ideal gas. Boyle's Law states that pressure is inversely proportional to volume. Charles's Law states that volume is in direct relation to temperature. Avogadro's Law shows that volume is in direct relation to the number of moles in the gas. The mathematical equation for the ideal gas law equation has been formulated by taking all the equations into account: PV=nRT Where P = pressure of the ideal gas V = volume of the ideal gas n = amount of ideal gas measured in moles R = universal gas constant and its value is 8.314 J.K-1mol-1 T = temperature

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm

Therefore, the diameter is 2.1 mm.

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Answer 4

Question

Definition Definition Law that is the combined form of Boyle's Law, Charles's Law, and Avogadro's Law. This law is obeyed by all ideal gas. Boyle's Law states that pressure is inversely proportional to volume. Charles's Law states that volume is in direct relation to temperature. Avogadro's Law shows that volume is in direct relation to the number of moles in the gas. The mathematical equation for the ideal gas law equation has been formulated by taking all the equations into account: PV=nRT Where P = pressure of the ideal gas V = volume of the ideal gas n = amount of ideal gas measured in moles R = universal gas constant and its value is 8.314 J.K-1mol-1 T = temperature

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm

Therefore, the diameter is 2.1 mm.

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Answer 5

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm

Therefore, the diameter is 2.1 mm.

Answer 6

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm

Therefore, the diameter is 2.1 mm.

Answer 7

Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Answer 8

Expert Solution & Answer

Answer to Problem 55P

The diameter is 2.1 mm.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Depth of lake is 80.0 m, temperature at the bottom of lake is 4°C, water temperature at the surface is 18°C, and the initial diameter of bubble is 1.00 mm.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

niR=nfR (I)

Here, ni is the initial number of moles, nf is the final number of moles, and R is the universal gas constant.

Write the equation for niR in terms of pressure, temperature, and the volume.

niR=PiViTi (II)

Here, Pi is the pressure at bottom, Vi is the volume at bottom, and Ti is the temperature at the bottom.

Write the equation for nfR in terms of pressure, temperature, and the volume.

nfR=PfVfTf (III)

Here, Pf is the pressure at top, Vf is the volume at top, and Tf is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

PiViTi=PfVfTf (IV)

Write the equation for Pi in terms of Pf.

Pi=Pf+ρgh (V)

Here, ρ is the density of water, g is the gravitational acceleration, and h is the depth of lake.

Write the equation for Vf.

Vf=16πdf2 (VI)

Here, df is the diameter of bubble at the top.

Write the equation for Vi.

Vi=16πdi2 (VII)

Here, di is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

(Pf+ρgh)(16πdi2)Ti=Pf(16πdf2)Tf

Rewrite the above relation in terms of df.

df=di((Pf+ρgd)TfPfTi)1/3

Conclusion:

Substitute 1.00 mm for di, 1.0 atm for Pf, 1.0×103 kg/m3 for ρ, 9.80 m/s2 for g, 80.0 m for h, 18°C for Tf, and 4°C for Ti in the above equation to find df.

df=(1.00 mm(10−3 m1 mm))((1.0 atm(1.013×105 Pa/atm1.0 atm)+(1.0×103 kg/m3)(9.80 m/s2)(80.0 m))(18°C)(273.15 K+18 K18°C)1.0 atm(1.013×105 Pa/atm1.0 atm)(4°C)(273.15 K+4 K4°C))1/3=(1.00 mm(10−3 m1 mm))(9.261)1/3=2.1×10−3 m(1 mm10−3 m)=2.1 mm