Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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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
Check Mark

Explanation of Solution

Given:

Load, F=110N

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

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

Temperature of air, T=0°C = 273 K

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

Net upward force on balloon, Fnet=30.0N

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=FbFwwHeFb=ρaVgwHe=ρHeVgFnet=ρaVgFwρHeVgV=Fnet+F+w(ρaρHe)gn=PVRTn=PRTFnet+F+w(ρaρHe)g

Density of air, ρa=1.293kg/m3

Density of helium, ρHe=0.179kg/m3

Substitute the values and solve:

  n=PRTFnet+F+w(ρaρHe)gn=(1atm)(30.0N+110N+50.0N)(8.21×102Latm/molK)(273K)(1.293kg/m30.179kg/m3)(9.81m/s2)(1L/103m-3)n776mol

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Load, F=110N

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

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

Temperature of air, T=0°C = 273 K

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

Net upward force on balloon, Fnet=30.0N

Formula Used:

Variation of pressure with altitude ( h ):

  Ph=PoeChC=0.13km-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=PoeChh=1Cln(PoPh)Ph=nRTVh=1Cln(PonRTV)h=1Cln(PoVnRT)

Substitute the values and solve:

  h=1Cln(PoVnRT)=10.13km-1ln(1.00atm×32×1L103m3(776mol)(8.206×102Latm/molK)(273K))=4.7km

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
Check Mark

Explanation of Solution

Given:

Load, F=110N

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

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

Temperature of air, T=0°C = 273 K

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

Net upward force on balloon, Fnet=30.0N

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=Fbwt0

  wt=F+w+wHewt=F+w+ρHeVgFb=ρa,hVgPhPo=ρa,hρaρa,h=PhPoρaFb=ρa,hVgFb=PhPoρaVgPh=PoeChFnet=eChVg(ρaρHe)Fw

Calculations:

Substitute the values and solve:

  Fnet=eChVg(ρaρHe)Fw=e(0.13km-1)(4.7km)(32.0m3)(9.81m/s2)(1.293kg/m30.179kg/m3)110N50N=30N

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
Check Mark

Explanation of Solution

Given:

Load, F=110N

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

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

Temperature of air, T=0°C = 273 K

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

Net upward force on balloon, Fnet=30.0N

Formula Used:

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

  Fnet=Fbwt=0

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

Calculations:

Substitute the values and solve:

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

Conclusion:

The maximum altitude attained by the balloon is 5.8km .

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Chapter 17 Solutions

Physics for Scientists and Engineers, Vol. 1

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