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The density of air at \(20^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) is \(1.205 \mathrm{~g} / \mathrm{L}\). If this air were compressed at the same temperature to equal the pressure at \(30.0 \mathrm{~m}\) 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 air density at 30m below sea level is higher than at the surface due to increased pressure.

Step by step solution

01

Identify Given Variables and Constants

We are given the density of air at the surface: \( \rho_1 = 1.205 \, \text{g/L} \). The initial pressure is \( P_1 = 1.00 \, \text{atm} \). The depth at which we need to calculate the new air density is \( h = 30.0 \, \text{m} \). The density of seawater is \( \rho_{\text{seawater}} = 1.025 \, \text{g/cm}^3 \) (convert it to g/L for consistency: \( \rho_{\text{seawater}} = 1025 \, \text{g/L} \)).
02

Calculate Pressure at 30m Depth

Use the formula for pressure at a certain depth: \[ P_{\text{depth}} = P_1 + \rho_{\text{seawater}} \cdot g \cdot h \]Where \( g = 9.81 \, \text{m/s}^2 \). Calculate the additional pressure due to 30m of seawater, convert units where necessary, and convert the result to atm to find \( P_2 \).
03

Apply Ideal Gas Law Concept to Relate Densities and Pressures

Since temperature remains constant, use the relation \( \frac{P_1}{\rho_1} = \frac{P_2}{\rho_2} \) from the Ideal Gas Law:\[ \rho_2 = \rho_1 \cdot \frac{P_2}{P_1} \]Substitute \( P_1 = 1.00 \, \text{atm} \), \( \rho_1 = 1.205 \, \text{g/L} \), and the previously calculated \( P_2 \) into the formula to find \( \rho_2 \).
04

Calculate the New Density \( \rho_2 \)

Use the formula from Step 3 to calculate \( \rho_2 \), the density at 30m below sea level. Perform the calculations for \( \rho_2 \) using the values you found for \( P_2 \) and the known values for \( \rho_1 \) and \( P_1 \). This will give the new density of air under the specified conditions.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a powerful tool for understanding how gases behave under varying conditions. At its core, it provides a relationship between pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the number of moles of gas (\(n\)). The law is usually expressed in the equation form: \(PV = nRT\). Here, \(R\) is the gas constant.

The Ideal Gas Law is particularly useful in scenarios involving changes in pressure or volume while keeping temperature constant, as in our exercise. When dealing with changes in these conditions, a derived formula often used is \( \frac{P_1}{\rho_1} = \frac{P_2}{\rho_2} \), where \(\rho\) represents the density of the gas.

This relationship shows how density is directly proportional to pressure when temperature remains unchanged. In simple terms, as the pressure increases, so does the density of the gas, given that the temperature is constant. This is essential in understanding many real-world phenomena, such as weather patterns and how gases behave in different environments.
Pressure and Depth
Understanding how pressure changes with depth is crucial, especially when analyzing environments under the sea. Pressure increases as we go deeper underwater due to the weight of the overlying water. The formula to calculate pressure at a certain depth is:

\[ P_{\text{depth}} = P_1 + \rho_{\text{seawater}} \cdot g \cdot h \]

Here, \(P_1\) represents the atmospheric pressure at the surface, \(\rho_{\text{seawater}}\) is the density of seawater, \(g\) is the acceleration due to gravity (approximated as \(9.81 \, \text{m/s}^2\)), and \(h\) is the depth.

In our exercise, the depth given is 30 meters. As we calculate the pressure at this depth, we consider the additional pressure exerted by the water column above. Converting this pressure to atmospheres allows us to find \(P_2\) used in the Ideal Gas Law equation. By doing so, we can then predict how other factors, like air density, change under these new pressure conditions.
Seawater Density
Seawater density plays a significant role in calculating how pressure varies with depth. It is denser than fresh water, meaning it exerts more pressure at the same depth. Seawater has an average density of about \(1.025 \, \text{g/cm}^3\), or \(1025 \, \text{g/L}\) when converted to more familiar density units for gases.

This density is important not just in calculating pressure changes but also affects buoyancy and the behavior of objects submerged in seawater. In our exercise, knowing the seawater density helped us in computing the increased pressure at 30 meters below sea level.

Understanding seawater density allows us to make predictions about marine environments and how things behave underwater. Its role in seawater pressure calculations is just a glimpse into its broader significance in oceanography and environmental science.

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Most popular questions from this chapter

Sodium hydrogen carbonate is also known as baking soda. When this compound is heated, it decomposes to sodium carbonate, carbon dioxide, and water vapor. Write the balanced equation for this reaction. What volume (in liters) of carbon dioxide gas at \(77^{\circ} \mathrm{C}\) and \(756 \mathrm{mmHg}\) will be produced from \(26.8 \mathrm{~g}\) of sodium hydrogen carbonate?

A \(1.000-\mathrm{g}\) sample of an unknown gas at \(0^{\circ} \mathrm{C}\) gives the following data: \(\begin{array}{lc}\boldsymbol{P}(\mathrm{atm}) & \boldsymbol{V}(\mathrm{L}) \\\ 0.2500 & 3.1908 \\ 0.5000 & 1.5928 \\ 0.7500 & 1.0601 \\ 1.0000 & 0.7930\end{array}\) Use these data to calculate the value of the molar mass at each of the given pressures from the ideal gas law (we will call this the "apparent molar mass" at this pressure). Plot the apparent molar masses against pressure and extrapolate to find the molar mass at zero pressure. Because the ideal gas law is most accurate at low pressures, this extrapolation will give an accurate value for the molar mass. What is the accurate molar mass?

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is an easily liquefied gaseous fuel. Calculate the density of butane gas at \(0.897 \mathrm{~atm}\) and \(24^{\circ} \mathrm{C}\). Give the answer in grams per liter.

An 18.6-mL volume of hydrochloric acid reacts completely with a solid sample of \(\mathrm{MgCO}_{3}\). The reaction is $$2 \mathrm{HCl}(a q)+\mathrm{MgCO}_{3}(s) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{MgCl}_{2}(a q)$$ The volume of \(\mathrm{CO}_{2}\) formed is \(159 \mathrm{~mL}\) at \(23^{\circ} \mathrm{C}\) and \(731 \mathrm{mmHg}\). What is the molarity of the \(\mathrm{HCl}\) solution?

What variables are needed to describe a gas that obeys the ideal gas law? What are the SI units for each variable?

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