Problem 2: Rate of Change of Soap Bubble Radius
1. Variable Relationships
Pressure: Inside pressure is atmospheric plus excess pressure due to surface tension ($4\sigma/r$ for a bubble with two surfaces).
$$ P = P_0 + \frac{4\sigma}{r} $$
Thickness ($h$): The volume of the soap film material ($V_s$) is constant.
$$ V_s = 4\pi r^2 h = 4\pi r_0^2 h_0 \implies h = h_0 \frac{r_0^2}{r^2} $$
Concentration Gradient: Using $n = P/kT$:
$$ n_i – n_o = \frac{1}{kT} (P – P_0) = \frac{1}{kT} \left( \frac{4\sigma}{r} \right) $$
2. Diffusion Equation
The rate of loss of molecules ($N$) is given by flux $q \times \text{Area}$.
$$ \frac{dN}{dt} = -q A = – \left( D \frac{n_i – n_o}{h} \right) 4\pi r^2 $$
Substitute expressions for $(n_i – n_o)$ and $h$:
$$ \frac{dN}{dt} = -D \frac{ \frac{4\sigma}{rkT} }{ h_0 \frac{r_0^2}{r^2} } 4\pi r^2 $$
$$ \frac{dN}{dt} = – \frac{16 \pi \sigma D r^3}{k T h_0 r_0^2} \quad \text{…(Eq 1)} $$
3. Differentiating the Gas Law
From $PV = NkT$, differentiate with respect to time:
$$ \frac{d}{dt} (PV) = \frac{dN}{dt} kT $$
$$ \frac{d}{dt} \left[ \left( P_0 + \frac{4\sigma}{r} \right) \left( \frac{4}{3}\pi r^3 \right) \right] = \frac{dN}{dt} kT $$
$$ \frac{4\pi}{3} \frac{d}{dt} (P_0 r^3 + 4\sigma r^2) = \frac{dN}{dt} kT $$
$$ \frac{4\pi}{3} (3 P_0 r^2 \dot{r} + 8\sigma r \dot{r}) = \frac{dN}{dt} kT $$
$$ 4\pi r \dot{r} \left( P_0 r + \frac{8\sigma}{3} \right) = \frac{dN}{dt} kT \quad \text{…(Eq 2)} $$
4. Final Solution
Equate Eq 1 and Eq 2 (solve for $\dot{r} = dr/dt$):
$$ 4\pi r \dot{r} \left( \frac{3 P_0 r + 8\sigma}{3} \right) = \left( – \frac{16 \pi \sigma D r^3}{k T h_0 r_0^2} \right) kT $$
$$ 4\pi r \dot{r} \left( \frac{3 P_0 r + 8\sigma}{3} \right) = – \frac{16 \pi \sigma D r^3}{h_0 r_0^2} $$
Cancel common terms ($4\pi r$):
$$ \dot{r} \left( \frac{3 P_0 r + 8\sigma}{3} \right) = – \frac{4 \sigma D r^2}{h_0 r_0^2} $$
$$ \frac{dr}{dt} = – \frac{12 \sigma D r^2}{h_0 r_0^2 (3 P_0 r + 8\sigma)} $$
Expression:
$$ \frac{dr}{dt} = – \frac{12 \sigma D}{h_0 r_0^2} \left( \frac{r^2}{3 P_0 r + 8\sigma} \right) $$
1. Variable Relationships
Pressure: Inside pressure is atmospheric plus excess pressure due to surface tension ($4\sigma/r$ for a bubble with two surfaces).
$$ P = P_0 + \frac{4\sigma}{r} $$Thickness ($h$): The volume of the soap film material ($V_s$) is constant. $$ V_s = 4\pi r^2 h = 4\pi r_0^2 h_0 \implies h = h_0 \frac{r_0^2}{r^2} $$
Concentration Gradient: Using $n = P/kT$:
$$ n_i – n_o = \frac{1}{kT} (P – P_0) = \frac{1}{kT} \left( \frac{4\sigma}{r} \right) $$2. Diffusion Equation
The rate of loss of molecules ($N$) is given by flux $q \times \text{Area}$.
$$ \frac{dN}{dt} = -q A = – \left( D \frac{n_i – n_o}{h} \right) 4\pi r^2 $$Substitute expressions for $(n_i – n_o)$ and $h$:
$$ \frac{dN}{dt} = -D \frac{ \frac{4\sigma}{rkT} }{ h_0 \frac{r_0^2}{r^2} } 4\pi r^2 $$ $$ \frac{dN}{dt} = – \frac{16 \pi \sigma D r^3}{k T h_0 r_0^2} \quad \text{…(Eq 1)} $$3. Differentiating the Gas Law
From $PV = NkT$, differentiate with respect to time:
$$ \frac{d}{dt} (PV) = \frac{dN}{dt} kT $$ $$ \frac{d}{dt} \left[ \left( P_0 + \frac{4\sigma}{r} \right) \left( \frac{4}{3}\pi r^3 \right) \right] = \frac{dN}{dt} kT $$ $$ \frac{4\pi}{3} \frac{d}{dt} (P_0 r^3 + 4\sigma r^2) = \frac{dN}{dt} kT $$ $$ \frac{4\pi}{3} (3 P_0 r^2 \dot{r} + 8\sigma r \dot{r}) = \frac{dN}{dt} kT $$ $$ 4\pi r \dot{r} \left( P_0 r + \frac{8\sigma}{3} \right) = \frac{dN}{dt} kT \quad \text{…(Eq 2)} $$4. Final Solution
Equate Eq 1 and Eq 2 (solve for $\dot{r} = dr/dt$):
$$ 4\pi r \dot{r} \left( \frac{3 P_0 r + 8\sigma}{3} \right) = \left( – \frac{16 \pi \sigma D r^3}{k T h_0 r_0^2} \right) kT $$ $$ 4\pi r \dot{r} \left( \frac{3 P_0 r + 8\sigma}{3} \right) = – \frac{16 \pi \sigma D r^3}{h_0 r_0^2} $$Cancel common terms ($4\pi r$):
$$ \dot{r} \left( \frac{3 P_0 r + 8\sigma}{3} \right) = – \frac{4 \sigma D r^2}{h_0 r_0^2} $$ $$ \frac{dr}{dt} = – \frac{12 \sigma D r^2}{h_0 r_0^2 (3 P_0 r + 8\sigma)} $$