/*! 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 49 A hydrogen balloon at Earth's su... [FREE SOLUTION] | 91影视

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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 m鲁.

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 鈮� \(14,995.75 \mathrm{m}^{3}\) 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

A bubble rises from the bottom of a lake of depth \(80.0 \mathrm{m},\) where the temperature is \(4^{\circ} \mathrm{C} .\) The water temperature at the surface is \(18^{\circ} \mathrm{C} .\) If the bubble's initial diameter is \(1.00 \mathrm{mm},\) what is its diameter when it reaches the surface? (Ignore the surface tension of water. Assume the bubble warms as it rises to the same temperature as the water and retains a spherical shape. Assume \(\left.P_{\mathrm{atm}}=1.0 \mathrm{atm} .\right)\)
What are the rms speeds of helium atoms, and nitrogen, hydrogen, and oxygen molecules at \(25^{\circ} \mathrm{C} ?\)
A scuba diver has an air tank with a volume of \(0.010 \mathrm{m}^{3}\) The air in the tank is initially at a pressure of \(1.0 \times 10^{7} \mathrm{Pa}\) Assuming that the diver breathes \(0.500 \mathrm{L} / \mathrm{s}\) of air, find how long the tank will last at depths of (a) \(2.0 \mathrm{m}\) and (b) $20.0 \mathrm{m} .\( (Make the same assumptions as in Example \)13.6 .)$
The driver from Practice Problem 13.3 fills his \(18.9-\mathrm{L}\) steel gasoline can in the morning when the temperature of the can and the gasoline is \(15.0^{\circ} \mathrm{C}\) and the pressure is 1.0 atm, but this time he remembers to replace the tightly fitting cap after filling the can. Assume that the can is completely full of gasoline (no air space) and that the cap does not leak. The temperature climbs to \(30.0^{\circ} \mathrm{C}\) Ignoring the expansion of the steel can, what would be the pressure of the heated gasoline? The bulk modulus for gasoline is $1.00 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$
The volume of air taken in by a warm-blooded vertebrate in the Andes mountains is \(210 \mathrm{L} /\) day at standard temperature and pressure (i.e., \(\left.0^{\circ} \mathrm{C} \text { and } 1 \text { atm }\right) .\) If the air in the lungs is at \(39^{\circ} \mathrm{C},\) under a pressure of $450 \mathrm{mm} \mathrm{Hg}$ and we assume that the vertebrate takes in an average volume of \(100 \mathrm{cm}^{3}\) per breath at the temperature and pressure of its lungs, how many breaths does this vertebrate take per day?
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