Question

The circuit shown in FIGURE P28.48 is inside a $15\mathrm{cm}$diameter balloon filled with helium that is kept at a constant pressure of $12\mathrm{atm}$. How long will it take the balloon’s diameter to increase to 16 cm?

Short answer

Time taken is$16\mathrm{s}.$

Step-by-step solution

Given information

We have given,

Diameter of the balloon =$15\mathrm{cm}$

Pressure =$1.2\mathrm{atm}$

We have to find the time taken by the balloon to change its diameter to$16\mathrm{cm}.$

Simplify

The equivalent resistance of resistances in parallel as shown in the figure

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{eq}1}}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R_{eq1}}=\frac{1}{200}+\frac{1}{300}=\frac{5}{600} \\ {R}_{eq1}=120\Omega \end{gathered}$$

Then the total resistance of the circuit will be

${\mathrm{R}}_{\mathrm{eq}}=120+280=400\Omega$

The total power dissipated by the circuit

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}=\frac{{\left(45\right)}^2}{400}=5.06$

then heat will be generated as

$$\begin{gathered} \mathrm{H}=\mathrm{Pt} \\ \mathrm{H}=5\mathrm{t} \end{gathered}$$

We know from the law of thermodynamics

localid="1649414991464" $$\begin{gathered} \mathrm{H}=\frac{3}{2}\mathrm{nRT}=5\mathrm{t} \\ 5\mathrm{t}=\frac{3}{2}\mathrm{P}({\mathrm{V}}_2-{\mathrm{V}}_1) \end{gathered}$$

Then,

$$\begin{gathered} {\mathrm{V}}_1-{\mathrm{V}}_2=\frac{4\mathrm{\pi }({(0.16)}^2-{(0.15)}^2)}{3} \\ {\mathrm{V}}_1-{\mathrm{V}}_2=0.14\times {10}^{-3}{\mathrm{m}}^3 \\ \mathrm{t}=\frac{3\times 12\times 0.14\times {10}^{-3}}{10} \\ \mathrm{t}=16\mathrm{s} \end{gathered}$$