Chapter 13, Problem 9STP

To determine

To Calculate: The new volume of the balloon.

Answer to Problem 9STP

  $V_2=1017.1\text{mL}$

Explanation of Solution

Given:

Volume of balloon at standard pressure, $V=125\text{mL}$

Standard pressure, $P=101.3\text{kPa}$

Depth at which balloon is placed in swimming pool, $h=1.27\text{m}$

Formula Used:

From Boyle’s law:

  $V_1{P}_1=V_2{P}_2$

Here, V represents volume and P represents Pressure.

Pressure of the balloon in swimming pool can be obtained by:

  $P=\rho gh$

Here, $\rho$ is the density of water, g is the gravitational acceleration and h is the depth.

Calculations:

Density of water, $\rho =\text{1000}{\text{kg/m}}^3$

Calculate the pressure of the balloon in swimming pool:

  $\begin{array}{c}P=\left(\text{1000}{\text{kg/m}}^3\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(1.27\text{m}\right)\\ =12.45\text{kPa}\end{array}$

Use Boyle’s law:

  $\begin{array}{l}{V}_1{P}_1=V_2{P}_2\\ V_2=\frac{V_1{P}_1}{P_2}\\ V_2=\frac{125\times {10}^{-6}{\text{m}}^{\text{3}}\times 101.3\text{kPa}}{12.45\text{kPa}}\\ V_2=1017.1\times {10}^{-6}{\text{m}}^{\text{3}}\\ V_2=1017.1\text{mL}\end{array}$

Conclusion:

Thus, the new volume of the balloon is $V_2=1017.1\text{mL}$ .