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Chapter 5: Problem 48

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 being cooled to \(1.00 \times 10^2 \mathrm{K}\) is approximately \(238.59 \mathrm{mL}\).

Step by step solution

01

Convert temperatures to Kelvin

Since we are dealing with temperatures in the context of a gas law, we need to work with temperatures in Kelvin. The initial temperature is given in Celsius, so we need to convert it to Kelvin using the formula: \(T(K) = T(^{\circ}C) + 273.15\) Now, let's convert the given initial temperature in Celsius to Kelvin: \(T_1(K) = 20.0^{\circ} C + 273.15 = 293.15 K\) The final temperature is already given in Kelvin as \(T_2 = 1.00 \times 10^{2} K\).
02

Use Charles's Law formula to find the final volume

Charles's Law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where \(V_1\) is the initial volume, \(T_1\) is the initial temperature, \(V_2\) is the final volume, and \(T_2\) is the final temperature. We have all the necessary values, so let's plug them into the formula: \(\frac{7.00 \times 10^2 \mathrm{mL}}{293.15 \mathrm{K}} = \frac{V_2}{1.00 \times 10^2 \mathrm{K}}\)
03

Solve for the final volume

To solve for the final volume, \(V_2\), we need to isolate it in the equation by multiplying both sides by the final temperature (\(1.00 \times 10^2 \mathrm{K}\)): \(V_2 = \frac{7.00 \times 10^2 \mathrm{mL} \times 1.00 \times 10^2 \mathrm{K}}{293.15 \mathrm{K}}\) Now calculate the value of \(V_2\): \(V_2 = \frac{7.00 \times 10^2 \mathrm{mL} \times 1.00 \times 10^2 \mathrm{K}}{293.15 \mathrm{K}} \approx 238.59 \mathrm{mL}\) The final volume of the balloon after being cooled to \(1.00 \times 10^2 \mathrm{K}\) is approximately \(238.59 \mathrm{mL}\).

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Most popular questions from this chapter

In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density $=1.0 \mathrm{g} / \mathrm{cm}^{3} )$ is allowed to take place in a bottle with a total volume of 825 \(\mathrm{mL}\) until 12\(\%\) by volume is ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( . Assuming that the \)\mathrm{CO}_{2}$ is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is $0.79 \mathrm{g} / \mathrm{cm}^{3} . )$

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