/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A bubble with a volume of \(1.00... [FREE SOLUTION] | 91Ó°ÊÓ

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A bubble with a volume of \(1.00 \mathrm{cm}^{3}\) forms at the bottom of a lake that is \(20.0 \mathrm{m}\) deep. The temperature at the bottom of the lake is \(10.0^{\circ} \mathrm{C} .\) The bubble rises to the surface where the water temperature is \(25.0^{\circ} \mathrm{C}\) Assume that the bubble is small enough that its temperature always matches that of its surroundings. What is the volume of the bubble just before it breaks the surface of the water? Ignore surface tension.

Short Answer

Expert verified
Answer: The final volume of the bubble just before it breaks the surface of the water is approximately \(1.053 \ cm^3\).

Step by step solution

01

State the Given Information and Charles' Law Equation

We are given the following information: Initial volume of the bubble \(= V_1 = 1.00 \ cm^3\) Initial temperature of the water at the bottom \(= T_1 = 10.0^{\circ} C\) Final temperature of the water at the surface \(= T_2 = 25.0^{\circ} C\) Our goal is to find the final volume of the bubble \((V_2)\) before it reaches the surface. To solve this problem, we'll use Charles' Law, which states that, for an ideal gas at constant pressure, the volume is directly proportional to the temperature (in Kelvin). The equation for Charles' Law is: \(V \propto T\) To find the relationship between the volume and temperature, we need to use the following equation, where k is a constant. \(V_1 / T_1 = V_2 / T_2\)
02

Convert Celsius Temperature to Kelvin

To work with the Charles' Law equation, we need to convert the given temperatures from Celsius to Kelvin. We can do this using the following formula: \(K = ^{\circ}C + 273.15\) Converting the temperatures, we get: \(T_1 = 10.0^{\circ} C + 273.15 = 283.15 K\) \(T_2 = 25.0^{\circ} C + 273.15 = 298.15 K\)
03

Solve for Final Volume \(V_2\) Using the Charles' Law Equation

Now that we have the temperatures in Kelvin, we can use the Charles' Law equation to solve for the final volume, \(V_2\): \(V_1 / T_1 = V_2 / T_2\) Rearranging to get \(V_2\): \(V_2 = (V_1 / T_1) \cdot T_2\) Plug in the given values of \(V_1\), \(T_1\), and \(T_2\): \(V_2 = (1.00 \ cm^3 / 283.15 K) \cdot 298.15 K\)
04

Calculate the Final Volume \(V_2\)

Solve for \(V_2\): \(V_2 = (1.00 \ cm^3) \cdot (298.15 K / 283.15 K)\) \(V_2 = 1.0529 \ cm^3\)
05

Report the Final Volume \(V_2\)

The final volume of the bubble just before it breaks the surface of the water is approximately \(1.053 \ cm^3\).

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Most popular questions from this chapter

These data are from a constant-volume gas thermometer experiment. The volume of the gas was kept constant, while the temperature was changed. The resulting pressure was measured. Plot the data on a pressure versus temperature diagram. Based on these data, estimate the value of absolute zero in Celsius. $$\begin{array}{cc} \hline T\left(^{\circ} \mathrm{C}\right) & P(\mathrm{atm}) \\\ 0 & 1.00 \\\ 20 & 1.07 \\\ 100 & 1.37 \\\ -33 & 0.88 \\\ -196 & 0.28 \\\ \hline \end{array}$$
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