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The density of air at \(20^{\circ} \mathrm{C}\) and 1.00 atm is \(1.205 \mathrm{~g} / \mathrm{L}\). If this air were compressed at the same temperature to equal the pressure at \(50.0 \mathrm{~m}\) below sea level, what would be its density? Assume the barometric pressure is constant at \(1.00 \mathrm{~atm} .\) The density of seawater is \(1.025 \mathrm{~g} / \mathrm{cm}^{3} .\)

Short Answer

Expert verified
The new density of the compressed air at 50m below sea level is 7.230 g/L.

Step by step solution

01

Calculate the Pressure at 50m Depth

The pressure underwater increases by 1 atm for every 10 meters of depth. Thus, at a depth of 50 meters, the pressure is the surface pressure of 1 atm plus the additional pressure due to the weight of the water column above it: \( 1.00 \, \text{atm} + \frac{50.0 \, \text{m}}{10 \, \text{m/atm}} = 6.00 \, \text{atm} \).
02

Apply Ideal Gas Law to Find New Density

Since the temperature is constant and using the ideal gas law, \( PV = nRT \). The density \( \rho \) changes with pressure as \( \rho_2 = \rho_1 \times \frac{P_2}{P_1} \). Substitute the pressures and initial density: \( \rho_2 = 1.205 \, \mathrm{g/L} \times \frac{6.00 \, \text{atm}}{1.00 \, \text{atm}} \).
03

Calculate the New Density

Calculate the new density: \( \rho_2 = 1.205 \, \mathrm{g/L} \times 6 = 7.230 \, \mathrm{g/L} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density
Density is a fundamental concept in physics and chemistry that describes how much mass is contained within a specific volume. It's often represented by the symbol \( \rho \) (the Greek letter rho). The formula for calculating density is:
\[ \rho = \frac{m}{V} \]Here, \( m \) is the mass of the substance, and \( V \) is its volume.

Understanding density helps in explaining why certain objects float while others sink. For gases, density can change under different conditions such as changes in pressure or temperature. In the context of the exercise, compressing air increases its density. This happens because more molecules are squeezed into the same space, increasing the mass per unit volume. Density is an essential factor when calculating other properties, like buoyancy, and in various practical applications such as engineering and meteorology.
Pressure
Pressure refers to the force exerted over an area. In physics, it's defined as force divided by area and is often measured in units like pascals (Pa) or atmospheres (atm). Pressure is a crucial concept when dealing with fluids and gases, as it can affect their behavior and properties.

In the solution provided for the exercise, pressure plays a significant role. The system describes how pressure increases linearly with depth when underwater, such as in the scenario of being 50 meters below sea level. The formula to determine this is based on the fact that the increase in depth results in an increase in pressure. As more water sits above the point of measurement, it adds to the atmospheric pressure, causing the total pressure to rise. Understanding pressure helps to predict how gases will react under different conditions, which is vital in understanding systems like weather patterns and in engineering applications.
Ideal Gas Law
The Ideal Gas Law is a fundamental principle used to describe the behavior of an ideal gas. The formula is expressed as follows:
\[ PV = nRT \]where:
  • \( P \) is the pressure
  • \( V \) is the volume
  • \( n \) is the number of moles of gas
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
In the exercise, the Ideal Gas Law is used to understand how density changes when pressure is altered, assuming temperature remains constant. When air is compressed, as when you dive deeper underwater, the volume decreases while the mass (number of molecules) remains the same, leading to an increase in density. This principle allows for the calculation of changing gas conditions in confined environments and is widely used across scientific disciplines to make predictions about how gases will respond to different factors like pressure and temperature.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the atmosphere. Standard atmospheric pressure at sea level is approximately 1 atm, or about 101,325 pascals (Pa). This pressure acts on all surfaces. It's what we typically measure when discussing pressure in everyday situations.

The exercise scenario assumes constant atmospheric pressure, providing a reference point for calculating the effects of additional pressures such as those first added by diving 50 meters below sea level. This localized increase in pressure is added on top of the standard atmospheric pressure. Understanding how atmospheric pressure interacts with other pressures is crucial for various applications. It affects weather systems, airplane cabin design, and the human body's ability to function at different altitudes.

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Most popular questions from this chapter

5.117 Liquid oxygen was first prepared by heating potassium chlorate, \(\mathrm{KClO}_{3},\) in a closed vessel to obtain oxygen at high pressure. The oxygen was cooled until it liquefied. $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ If \(171 \mathrm{~g}\) of potassium chlorate reacts in a \(2.70-\mathrm{L}\) vessel, which was initially evacuated, what pressure of oxygen will be attained when the temperature is finally cooled to \(25^{\circ} \mathrm{C} ?\) Use the preceding chemical equation and ignore the volume of solid product.

Formic acid, \(\mathrm{HCHO}_{2},\) is a convenient source of small quantities of carbon monoxide. When warmed with sulfuric acid, formic acid decomposes to give CO gas. $$ \mathrm{HCHO}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}(g) $$ If \(3.85 \mathrm{~L}\) of carbon monoxide was collected over water at \(25^{\circ} \mathrm{C}\) and \(689 \mathrm{mmHg}\), how many grams of formic acid were consumed?

5.161 Power plants driven by fossil-fuel combustion generate substantial greenhouse gases (e.g., \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) as well as gases that contribute to poor air quality (e.g., \(\mathrm{SO}_{2}\) ). To evaluate exhaust emissions for regulation purposes, the generally inert nitrogen supplied with air must be included in the balanced reactions; it is further assumed that air is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a volume/volume ratio of \(3.76: 1.00,\) respectively. In addition to the water produced, the fuel's \(\mathrm{C}\) and \(\mathrm{S}\) are converted to \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) at the stack. Given these stipulations, answer the following for a power plant running on a fossil fuel having the formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\). Including the \(\mathrm{N}_{2}\) supplied in air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the product gases). Except for \(\mathrm{N}_{2},\) use only integer coefficients. (See problem \(3.141 .\) ) What gas law is used to effectively convert the given \(\mathrm{N}_{2} / \mathrm{O}_{2}\) volume ratio to a molar ratio when deriving the balanced combustion equation in (a)? (i) Boyle's law (ii) Charles's law (iii) Avogadro's law (iv) ideal gas law c) Assuming the product water condenses, use the result from (b) to determine the stack gas composition on a "dry" basis by calculating the volume/volume percentages for \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{SO}_{2}\) Assuming the product water remains as vapor, repeat (c) on a "wet" basis.Assuming the product water remains as vapor, repeat (c) on a "wet" basis. Some fuels are cheaper than others, particularly those with higher sulfur contents that lead to poorer air quality. In port, when faced with ever-increasing air quality regulations, large ships operate power plants on more costly, higher-grade fuels; with regulations nonexistent at sea, ships' officers make the cost-saving switch to lower-grade fuels. Suppose two fuels are available, one having the general formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and the other \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S} .\) Based on air quality concerns alone, which fuel is more likely to be used when idling in port? Support your answer to (d) by calculating the mass ratio of \(\mathrm{SO}_{2}\) produced to fuel burned for \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S},\) assuming \(120 \%\) stoichiometric combustion.

Nitric acid is produced from nitrogen monoxide, NO, which in turn is prepared from ammonia by the Ostwald process: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What volume of oxygen at \(35^{\circ} \mathrm{C}\) and 2.15 atm is needed to produce \(100.0 \mathrm{~g}\) of nitrogen monoxide?

Magnesium metal reacts with hydrochloric acid to produce hydrogen gas, \(\mathrm{H}_{2}\). $$ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Calculate the volume (in liters) of hydrogen produced at \(33^{\circ} \mathrm{C}\) and \(665 \mathrm{mmHg}\) from \(0.0840 \mathrm{~mol} \mathrm{Mg}\) and excess \(\mathrm{HCl}\).

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