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Chapter 13: Problem 49

A hydrogen balloon at Earth's surface has a volume of \(5.0 \mathrm{m}^{3}\) on a day when the temperature is \(27^{\circ} \mathrm{C}\) and the pressure is \(1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2} .\) The balloon rises and expands as the pressure drops. What would the volume of the same number of moles of hydrogen be at an altitude of \(40 \mathrm{km}\) where the pressure is \(0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) and the temperature is \(-13^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The volume of the hydrogen balloon at 40 km altitude is approximately 14,995.75 m³.

Step by step solution

01

Convert temperature to Kelvin

Before using the combined gas law formula, we need to convert the temperatures to Kelvin. To do so, we add 273.15 to the Celsius temperature. Initial Temperature (T1) = \(27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K}\) Final Temperature (T2) = \(-13^{\circ} \mathrm{C} + 273.15 = 260.15 \mathrm{K}\)
02

Plug in the values into the combined gas law formula

We will now input the given values and the temperatures in Kelvin into the combined gas law formula: (P1 * V1) / T1 = (P2 * V2) / T2 where P1 = \(1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) V1 = \(5.0 \mathrm{m}^{3}\) T1 = \(300.15 \mathrm{K}\) P2 = \(0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) T2 = \(260.15 \mathrm{K}\)
03

Solve for V2

Now, we will solve for V2 (the final volume at 40 km altitude). Rearrange the combined gas law formula to solve for V2: V2 = (P1 * V1 * T2) / (P2 * T1) Plug in the values: V2 = (\((1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2})\) * \((5.0 \mathrm{m}^{3})\) * \((260.15 \mathrm{K})\)) / (\((0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2})\) * \((300.15 \mathrm{K})\)) Now, calculate V2: V2 ≈ \(14,995.75 \mathrm{m}^{3}\) So, the volume of the hydrogen balloon at 40 km altitude is approximately \(14,995.75 \mathrm{m}^{3}\).

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Most popular questions from this chapter

Agnes Pockels \((1862-1935)\) was able to determine Avogadro's number using only a few household chemicals, in particular oleic acid, whose formula is \(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\) (a) What is the molar mass of this acid? (b) The mass of one drop of oleic acid is \(2.3 \times 10^{-5} \mathrm{g}\) and the volume is $2.6 \times 10^{-5} \mathrm{cm}^{3} .$ How many moles of oleic acid are there in one drop? (c) Now all Pockels needed was to find the number of molecules of oleic acid. Luckily, when oleic acid is spread out on water, it lines up in a layer one molecule thick. If the base of the molecule of oleic acid is a square of side \(d\), the height of the molecule is known to be \(7 d .\) Pockels spread out one drop of oleic acid on some water, and measured the area to be \(70.0 \mathrm{cm}^{2}\) Using the volume and the area of oleic acid, what is \(d ?\) (d) If we assume that this film is one molecule thick, how many molecules of oleic acid are there in the drop? (e) What value does this give you for Avogadro's number?
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