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Chapter 5: Problem 42

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{~mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\). The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{~K}\). What is the final volume of the balloon?

Short Answer

Expert verified
The final volume of the balloon after cooling is approximately \(238.8 \mathrm{~mL}\).

Step by step solution

01

Write down the Charles' Law formula.

Charles' Law states that at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. The formula for Charles' Law is: \(V_1/T_1 = V_2/T_2\) where, - \(V_1\) is the initial volume of the gas - \(T_1\) is the initial temperature in Kelvin - \(V_2\) is the final volume of the gas - \(T_2\) is the final temperature in Kelvin
02

Convert the given temperatures to Kelvin.

The initial temperature is given in Celsius, and we need to convert it to Kelvin using the formula: \(T(K) = T(^\circ C) + 273.15\) Initial temperature in Kelvin = \(20.0 + 273.15 = 293.15\ \mathrm{K}\) The final temperature is already given in Kelvin.
03

Substitute the given values into the Charles' Law formula.

Now, we have all the necessary values, and we can substitute them into the Charles' Law formula: \(\frac{7.00 \times 10^{2} \mathrm{~mL}}{293.15 \mathrm{~K}} = \frac{V_2}{1.00 \times 10^{2} \mathrm{~K}}\)
04

Solve for the final volume \(V_2\).

To find the final volume, we can cross-multiply and solve for \(V_2\): \(V_2 = \frac{7.00 \times 10^{2} \mathrm{~mL} \times 1.00 \times 10^{2} \mathrm{~K}}{293.15 \mathrm{~K}}\)
05

Calculate the final volume.

Now, compute the final volume: \(V_2 = \frac{7.00 \times 10^{2} \mathrm{~mL} \times 1.00 \times 10^{2} \mathrm{~K}}{293.15 \mathrm{~K}} = 238.8 \mathrm{~mL}\) The final volume of the balloon after cooling is approximately \(238.8 \mathrm{~mL}\).

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Most popular questions from this chapter

A sealed balloon is filled with \(1.00 \mathrm{~L}\) helium at \(23^{\circ} \mathrm{C}\) and \(1.00\) atm. The balloon rises to a point in the atmosphere where the pressure is 220 . torr and the temperature is \(-31^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from \(1.00\) atm to a pressure of 220 . torr?

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