Question

A balloon contains 14.0 L of air at 760 torr. What will be the volume of the balloon when it is taken to a depth of 10ft in a swimming pool? Assume that the temperature of the air and water are equal.
(Density: $Hg = 13.6\dfrac{g}{{mL}}$ )
A.11.0
B.11.3
C.10
D.10.8

Answer 1

To solve this question, we need to first establish a relation between the volume and the pressure of a gas. Then we must find the pressure at the final reading from the given data. Then, using all the components of data available to us, we can calculate the final volume of the balloon.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
The data that has been given to us can be represented as follows:
1. ${P_1} = 760$ torr = atm
2. ${V_1} = 140L$
3.Depth = 10 ft = 3.048 m
We know that $P = h\rho g$
The second reading is taken at a depth of 10 ft below water. Hence, while calculating the final volume, along with the atmospheric pressure, an additional pressure because of the water will be applied. Hence, the value of pressure at the final reading can be calculated as:
${P_2} =$ (atmospheric pressure) + (pressure due to water)
$\Rightarrow$ ${P_2} = \left( {1atm} \right) + (h\rho g)$
$= \left( 1 \right) + \left( {3.048} \right)\left( {1000} \right)\left( {9.81} \right)$
=1.295 atm
Substituting these values in Boyle’s equation we get:
${P_1}{V_1} = {P_2}{V_2}$
$\Rightarrow$ $(1)(14.0) = (1.295)({V_2})$
$\Rightarrow$ ${V_2} = 10.8L$

Hence, Option D is the correct option

Note: According to Boyle’ Law, for a given gas at constant temperature, the pressure of a given quantity of the gas is inversely proportional to the volume of the gas. This inverse relation can be applied in a gas at two different observations as:
${P_1}{V_1} = {P_2}{V_2}$
Where ${P_1},\,{V_1}$ are the values of the pressure and volume at the initial observation, while ${P_2},\,{V_2}$ are the values of the pressure and volume at the final observation.