The number of moles of Helium gas contained in the balloon.

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Answer 1

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Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

(d)

To determine

The maximum altitude attained by the balloon.

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

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Answer 2

Question

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Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

(d)

To determine

The maximum altitude attained by the balloon.

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

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Answer 3

Question

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Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

(d)

To determine

The maximum altitude attained by the balloon.

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

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Answer 4

Question

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Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

(d)

To determine

The maximum altitude attained by the balloon.

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

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Answer 5

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Answer 6

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Answer 7

Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

(d)

To determine

The maximum altitude attained by the balloon.

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

Answer 8

Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

Answer 9

Chapter 17, Problem 58P

(a)

To determine

To Calculate: The number of moles of Helium gas contained in the balloon.

Answer 10

(a)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Buoyant force can be obtained by:

$Fb=ρVg$

here, $ρ$ is the density of the fluid, V is the volume of the displaced fluid and g is the acceleration due to gravity.

Calculations:

$Fnet=Fb−F−w−wHeFb=ρaVgwHe=ρHeVg⇒Fnet=ρaVg−F−w−ρHeVg⇒V=Fnet+F+w(ρa−ρHe)gn=PVRT⇒n=PRTFnet+F+w(ρa−ρHe)g$

Density of air, $ρa=1.293 kg/m3$

Density of helium, $ρHe=0.179 kg/m3$

Substitute the values and solve:

$n=PRTFnet+F+w(ρa−ρHe)gn=(1atm)(30.0N+110 N+50.0 N)(8.21×10−2L⋅atm/mol⋅K)(273K)(1.293 kg/m3−0.179 kg/m3)(9.81m/s2)(1L/10−3m-3)n≈776 mol$

Conclusion:

Thus, the number of moles of Helium gas contained in the balloon is $776 mol$ .

Answer 11

(a)

Expert Solution

Answer 12

(a)

Expert Solution

Answer 13

Expert Solution

Answer 14

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

Answer 15

(b)

To determine

To Find:The altitudeat which the balloon would be fully inflated.

Answer 16

(b)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

Variation of pressure with altitude ( h ):

$Ph=Poe−ChC=0.13 km-1$

From ideal gas law:

$PV=nRT$

Here, P is the pressure, V is the volume, R is the gas constant, n is the number of moles and T is the temperature.

Calculations:

$Ph=Poe−Ch⇒h=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)⇒h=1Cln(PoVnRT)$

Substitute the values and solve:

$h=1Cln(PoVnRT)=10.13 km-1ln(1.00 atm×32 ×1 L10−3m3(776 mol)(8.206×10−2L⋅atm/mol⋅K)(273 K))=4.7 km$

Conclusion:

Thus, the altitude at which the balloon would be fully inflated is 4.7 km.

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

Answer 21

(c)

To determine

Whether the balloon would ever reach the altitude at which it is fully inflated.

Answer 22

(c)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

In order to reach the altitude at which the balloon is fully inflated, the buoyant force $(Fb)$ must be at least equal to the total weight of the balloon $wt$ or should be greater than it.

$Fnet=Fb−wt≥0$

$wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρa⇒ρa,h=PhPoρaFb=ρa,hVg⇒Fb=PhPoρaVgPh=Poe−ChFnet=e−ChVg(ρa−ρHe)−F−w$

Calculations:

Substitute the values and solve:

$Fnet=e−ChVg(ρa−ρHe)−F−w=e−(0.13km-1)(4.7km)(32.0 m3)(9.81 m/s2)(1.293kg/m3−0.179 kg/m3)−110 N−50 N=30 N$

Because $Fnet>0$ , the balloon would rise higher than the altitude where it is fully inflated.

Conclusion:

Yes, the balloon would ever reach the altitude at which it is fully inflated.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

(d)

To determine

The maximum altitude attained by the balloon.

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

Answer 27

(d)

To determine

The maximum altitude attained by the balloon.

Answer 28

(d)

Expert Solution

Explanation of Solution

Given:

Load, $F=110 N$

Weight of the balloon’s envelope, $w=50.0 N$

Volume of the balloon when it is fully inflated, $V=32.0 m3$

Temperature of air, $T=0°C = 273 K$

Atmospheric pressure, $P=1 atm =1.01×105Pa$

Net upward force on balloon, $Fnet=30.0 N$

Formula Used:

The balloon will rise till the net force on it is zero.

$Fnet=Fb−wt=0$

$h=1Clnρaρa,hρa,h=FbVgh=1Cln[ρaFbVg]h=1Cln[VgρaFb]$

Calculations:

Substitute the values and solve:

$h=10.13 km-1ln[(32.0 m3)(9.81m/s2)(1.293 kg/m3)190.5 N]h=5.8 km$

Conclusion:

The maximum altitude attained by the balloon is $5.8 km$ .

Answer 29

(d)

Expert Solution

Answer 30

(d)

Expert Solution

Answer 31

Expert Solution

Answer 32

Ch. 17 - Prob. 1P Ch. 17 - Prob. 2P Ch. 17 - Prob. 3P Ch. 17 - Prob. 4P Ch. 17 - Prob. 5P Ch. 17 - Prob. 6P Ch. 17 - Prob. 7P Ch. 17 - Prob. 8P Ch. 17 - Prob. 9P Ch. 17 - Prob. 10P

The number of moles of Helium gas contained in the balloon.

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To Calculate: The number of moles of Helium gas contained in the balloon.

Explanation of Solution

To Find:The altitudeat which the balloon would be fully inflated.

Explanation of Solution

Whether the balloon would ever reach the altitude at which it is fully inflated.

Explanation of Solution

The maximum altitude attained by the balloon.

Explanation of Solution

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