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Chapter 5: Problem 70

A hot-air balloon is filled with air to a volume of \(4.00 \times\) $10^{3} \mathrm{m}^{3}\( at 745 torr and \)21^{\circ} \mathrm{C}$ . The air in the balloon is then heated to \(62^{\circ} \mathrm{C},\) causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{m}^{3} .\) What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

Short Answer

Expert verified
The ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon is approximately 0.955. This means that there are slightly fewer moles of air in the heated balloon compared to the original number of moles in the balloon.

Step by step solution

01

Convert temperatures to Kelvin

Convert the given temperatures in Celsius to Kelvin by adding 273.15 to each temperature. Initial temperature: \(T_1 = 21^{\circ} \mathrm{C} + 273.15 = 294.15 \mathrm{K}\) Final temperature: \(T_2 = 62^{\circ} \mathrm{C} + 273.15 = 335.15 \mathrm{K}\)
02

Calculate the initial and final number of moles using the ideal gas law

Using the ideal gas law, PV = nRT, we can rearrange it to solve for the number of moles (n) in each case: \(n_1 = \frac{P_1V_1}{R_1T_1}\) and \(n_2 = \frac{P_2V_2}{R_2T_2}\) Since the pressure remains constant, we have \(P_1 = P_2\) and since we are dealing with the same gas, we can assume that \(R_1 = R_2\). Therefore, the equation simplifies to: \(n_1 = \frac{V_1}{RT_1}\) and \(n_2 = \frac{V_2}{RT_2}\) We know the initial volume, final volume, and temperatures in Kelvin, so we can calculate the initial and final number of moles.
03

Calculate the ratio of moles

Now that we have expressions for \(n_1\) and \(n_2\), we can calculate the ratio of moles: \(\frac{n_2}{n_1} = \frac{\frac{V_2}{RT_2}}{\frac{V_1}{RT_1}}\) With the R's canceling out, we are left with: \(\frac{n_2}{n_1} = \frac{V_2T_1}{V_1T_2}\) Substitute the given values of initial volume \(V_1 = 4.00 \times 10^3 \mathrm{m}^3\), final volume \(V_2 = 4.20 \times 10^3 \mathrm{m}^3\), initial temperature \(T_1 = 294.15 \mathrm{K}\), and final temperature \(T_2 = 335.15 \mathrm{K}\): \(\frac{n_2}{n_1} = \frac{(4.20 \times 10^3 \mathrm{m}^3)(294.15 \mathrm{K})}{(4.00 \times 10^3 \mathrm{m}^3)(335.15 \mathrm{K})}\)
04

Compute the result

Now we can calculate the value of the ratio: \(\frac{n_2}{n_1} = 0.955\) Thus, the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon is approximately 0.955. This means that there are slightly fewer moles of air in the heated balloon compared to the original number of moles in the balloon.

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