IPhO 2004 · Theory problem 2

Rising Balloon

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A rubber balloon filled with helium gas goes up high into the sky where the pressure and temperature decrease with height. In the following questions, assume that the shape of the balloon remains spherical regardless of the payload, and neglect the payload volume. Also assume that the temperature of the helium gas inside of the balloon is always the same as that of the ambient air, and treat all gases as ideal gases. The universal gas constant is \(R=8.31\ \mathrm{J/mol\cdot K}\) and the molar masses of helium and air are \(M_H=4.00\times10^{-3}\ \mathrm{kg/mol}\) and \(M_A=28.9\times10^{-3}\ \mathrm{kg/mol}\), respectively. The gravitational acceleration is \(g=9.8\ \mathrm{m/s^2}\).

[Part A ]

(a) [1.5 points] Let the pressure of the ambient air be \(P\) and the temperature be \(T\). The pressure inside of the balloon is higher than that of outside due to the surface tension of the balloon. The balloon contains \(n\) moles of helium gas and the pressure inside is \(P+\Delta P\). Find the buoyant force \(F_B\) acting on the balloon as a function of \(P\) and \(\Delta P\).

(b) [2 points] On one particular summer day in Korea, the air temperature \(T\) at the height \(z\) from the sea level was found to be \(T(z)=T_0(1-z/z_0)\) in the range of \(0<z<15\ \mathrm{km}\) with \(z_0=49\ \mathrm{km}\) and \(T_0=303\ \mathrm{K}\). The pressure and density at the sea level were \(P_0=1.0\ \mathrm{atm}=1.01\times10^5\ \mathrm{Pa}\) and \(\rho_0=1.16\ \mathrm{kg/m^3}\), respectively. For this height range, the pressure takes the form

\[ P(z)=P_0(1-z/z_0)^\eta. \tag{2.1} \]

Express \(\eta\) in terms of \(z_0\), \(\rho_0\), \(P_0\), and \(g\), and find its numerical value to the two significant digits. Treat the gravitational acceleration as a constant, independent of height.

[Part B ]

When a rubber balloon of spherical shape with un-stretched radius \(r_0\) is inflated to a sphere of radius \(r\) \((\ge r_0)\), the balloon surface contains extra elastic energy due to the stretching. In a simplistic theory, the elastic energy at constant temperature \(T\) can be expressed by

\[ U=4\pi r_0^2\kappa RT \left(2\lambda^2+\frac{1}{\lambda^4}-3\right) \tag{2.2} \]

where \(\lambda\equiv r/r_0\) \((\ge1)\) is the size-inflation ratio and \(\kappa\) is a constant in units of \(\mathrm{mol/m^2}\).

(c) [2 points] Express \(\Delta P\) in terms of parameters given in Eq. (2.2), and sketch \(\Delta P\) as a function of \(\lambda=r/r_0\).

(d) [1.5 points] The constant \(\kappa\) can be determined from the amount of the gas needed to inflate the balloon. At \(T_0=303\ \mathrm{K}\) and \(P_0=1.0\ \mathrm{atm}=1.01\times10^5\ \mathrm{Pa}\), an un-stretched balloon \((\lambda=1)\) contains \(n_0=12.5\) moles of helium. It takes \(n=3.6n_0=45\) moles in total to inflate the balloon to \(\lambda=1.5\) at the same \(T_0\) and \(P_0\). Express the balloon parameter \(a\), defined as \(a=\kappa/\kappa_0\), in terms of \(n\), \(n_0\), and \(\lambda\), where \(\kappa_0\equiv r_0P_0/(4RT_0)\). Evaluate \(a\) to the two significant digits.

[Part C]

A balloon is prepared as in (d) at the sea level (inflated to \(\lambda=1.5\) with \(n=3.6n_0=45\) moles of helium gas at \(T_0=303\ \mathrm{K}\) and \(P_0=1\ \mathrm{atm}=1.01\times10^5\ \mathrm{Pa}\)). The total mass including gas, balloon itself, and other payloads is \(M_T=1.12\ \mathrm{kg}\). Now let the balloon rise from the sea level.

(e) [3 points] Suppose that the balloon eventually stops at the height \(z_f\) where the buoyant force balances the total weight. Find \(z_f\) and the inflation ratio \(\lambda_f\) at that height. Give the answers in two significant digits. Assume there are no drift effect and no gas leakage during the upward flight.

Official English text of the problem.

Answers

  • (a) The upward buoyant force is \[ \boxed{F_B=nM_Ag\,\frac{P}{P+\Delta P}}. \]

  • (b) \[ \boxed{\eta=\frac{\rho_0gz_0}{P_0}=5.5}. \]

  • (c) \[ \boxed{\Delta P=\frac{4\kappa RT}{r_0} \left(\lambda^{-1}-\lambda^{-7}\right)}. \] The curve starts at \((1,0)\), reaches its single maximum at \(\lambda=7^{1/6}\approx1.38\), and then decreases toward zero from above.

  • (d) \[ \boxed{a=\frac{n/n_0-\lambda^3}{\lambda^2-\lambda^{-4}}=0.11}. \]

  • (e) \[ \boxed{z_f=1.1\times10^4\ \mathrm{m}=11\ \mathrm{km}, \qquad \lambda_f=2.1}. \] The unrounded results are \(z_f=11.095\ \mathrm{km}\) and \(\lambda_f=2.14594\); using the rounded values \(\eta=5.5\) and \(a=0.11\) instead gives the same answers to two significant digits.

Solution

(a)

By Archimedes’ principle, the buoyant force equals the weight of the displaced air. Treating ambient density as uniform over the balloon’s dimensions,

\[ F_B=\rho_A Vg. \]

The ideal-gas equations for the surrounding air and the helium give

\[ \rho_A=\frac{PM_A}{RT}, \qquad V=\frac{nRT}{P+\Delta P}. \]

The common temperature therefore cancels:

\[ \boxed{F_B=nM_Ag\,\frac{P}{P+\Delta P}}. \]

This is the buoyant force alone. The weights of the helium, envelope, and payload enter separately in the total force balance.

(b)

The atmosphere is in hydrostatic equilibrium:

\[ \frac{dP}{dz}=-\rho(z)g. \]

For ideal air of constant composition,

\[ \frac{\rho(z)}{\rho_0} =\frac{P(z)}{P_0}\frac{T_0}{T(z)}. \]

Using the prescribed temperature and pressure profiles,

\[ \rho(z)=\rho_0 \left(1-\frac{z}{z_0}\right)^{\eta-1}. \]

Differentiating Eq. (2.1) and substituting into the hydrostatic equation gives

\[ -\frac{\eta P_0}{z_0} \left(1-\frac{z}{z_0}\right)^{\eta-1} = -\rho_0g \left(1-\frac{z}{z_0}\right)^{\eta-1}. \]

Consequently,

\[ \boxed{\eta=\frac{\rho_0gz_0}{P_0}}. \]

Numerically,

\[ \eta= \frac{(1.16)(9.8)(49\times10^3)} {1.01\times10^5} =5.51517\ldots, \]

so, to two significant digits,

\[ \boxed{\eta=5.5}. \]

(c)

For a quasistatic expansion at fixed temperature, the work done by the excess internal pressure supplies the increase in the modeled elastic energy:

\[ \Delta P\,dV=dU. \]

The balloon volume is

\[ V=\frac{4\pi}{3}r_0^3\lambda^3, \qquad \frac{dV}{d\lambda}=4\pi r_0^3\lambda^2. \]

From Eq. (2.2),

\[ \left(\frac{\partial U}{\partial\lambda}\right)_T =16\pi r_0^2\kappa RT \left(\lambda-\lambda^{-5}\right). \]

Thus

\[ \boxed{ \Delta P =\frac{(\partial U/\partial\lambda)_T}{dV/d\lambda} =\frac{4\kappa RT}{r_0} \left(\lambda^{-1}-\lambda^{-7}\right) }. \]

To specify the sketch, define the pressure scale

\[ C=\frac{4\kappa RT}{r_0}. \]

Then

\[ \frac{d\Delta P}{d\lambda} =C\lambda^{-8}(7-\lambda^6). \]

The curve has the following features:

  • Its physical domain is \(\lambda\ge1\), and it starts at \(\Delta P=0\) when \(\lambda=1\).
  • It rises with positive slope until \[ \lambda_{\max}=7^{1/6}=1.38309\ldots. \]
  • Its maximum value is \[ \Delta P_{\max} =C\frac{6}{7^{7/6}} \approx0.620\,C. \]
  • After the maximum, it decreases monotonically, remaining positive. At large inflation, \[ \Delta P\sim\frac{C}{\lambda}\longrightarrow0^+, \] so the horizontal axis is an asymptote.

(d)

Let

\[ V_0=\frac{4\pi}{3}r_0^3. \]

At \(\lambda=1\), part (c) gives \(\Delta P=0\). Hence the un-stretched balloon satisfies

\[ P_0V_0=n_0RT_0. \]

Since

\[ a=\frac{\kappa}{\kappa_0} =\frac{4\kappa RT_0}{r_0P_0}, \]

the excess pressure at temperature \(T_0\) is

\[ \frac{\Delta P}{P_0} =a\left(\lambda^{-1}-\lambda^{-7}\right). \]

The inflated balloon satisfies

\[ (P_0+\Delta P)V_0\lambda^3=nRT_0. \]

Dividing by \(P_0V_0=n_0RT_0\),

\[ \frac{n}{n_0} =\lambda^3\left[ 1+a\left(\lambda^{-1}-\lambda^{-7}\right) \right] =\lambda^3+a\left(\lambda^2-\lambda^{-4}\right). \]

Therefore,

\[ \boxed{ a=\frac{n/n_0-\lambda^3}{\lambda^2-\lambda^{-4}} }. \]

Substitution gives

\[ a=\frac{3.6-1.5^3}{1.5^2-1.5^{-4}} =0.109624\ldots, \]

and hence

\[ \boxed{a=0.11}. \]

(e)

Define the dimensionless temperature

\[ t=\frac{T}{T_0}=1-\frac{z}{z_0}. \]

Then

\[ P=P_0t^\eta, \qquad \Delta P=aP_0t \left(\lambda^{-1}-\lambda^{-7}\right). \]

Because no helium escapes, \(n\) remains constant. Its ideal-gas equation is

\[ \left[ P_0t^\eta+aP_0t \left(\lambda^{-1}-\lambda^{-7}\right) \right]V_0\lambda^3 =nRT_0t. \]

Using \(P_0V_0=n_0RT_0\) and dividing by \(t\),

\[ \frac{n}{n_0} =t^{\eta-1}\lambda^3 +a\left(\lambda^2-\lambda^{-4}\right). \tag{1} \]

The displaced air mass follows from its ideal-gas density:

\[ \rho_A V =\frac{M_AP}{RT}V_0\lambda^3 =n_0M_A\,t^{\eta-1}\lambda^3. \]

At the final height, buoyancy balances the entire weight:

\[ n_0M_Ag\,t_f^{\eta-1}\lambda_f^3=M_Tg, \]

so

\[ t_f^{\eta-1}\lambda_f^3 =\frac{M_T}{n_0M_A}. \tag{2} \]

The helium’s weight is already included in \(M_Tg\).

Eliminating the atmospheric factor between (1) and (2) gives an equation for the final inflation alone:

\[ a\left(\lambda_f^2-\lambda_f^{-4}\right) =\frac{n}{n_0}-\frac{M_T}{n_0M_A}. \]

Keeping unrounded intermediate values,

\[ \frac{M_T}{n_0M_A} =\frac{1.12}{(12.5)(28.9\times10^{-3})} =3.100346\ldots, \]

and therefore

\[ \lambda_f^2-\lambda_f^{-4} =\frac{3.6-3.100346\ldots}{0.109624\ldots} =4.557886\ldots. \]

The left side increases strictly for \(\lambda_f\ge1\), because its derivative is \(2\lambda_f+4\lambda_f^{-5}>0\). Thus there is a unique physical root:

\[ \lambda_f=2.145936\ldots. \]

Equation (2) now gives

\[ t_f= \left( \frac{M_T}{n_0M_A\lambda_f^3} \right)^{1/(\eta-1)} =0.773571\ldots. \]

Consequently,

\[ z_f=z_0(1-t_f) =(49\ \mathrm{km})(1-0.773571\ldots) =11.0950\ldots\ \mathrm{km}. \]

To two significant digits,

\[ \boxed{\lambda_f=2.1,\qquad z_f=11\ \mathrm{km}}. \]

The final height lies within the specified atmospheric range. The balloon also has positive initial lift: at sea level, its displaced air mass is

\[ n_0M_A(1.5)^3=1.2192\ldots\ \mathrm{kg}>M_T. \]

The solution was checked against the official solution.