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Chapter 18: Problem 1

A \(20.0 \mathrm{~L}\) tank contains \(4.86 \times 10^{-4} \mathrm{~kg}\) of helium at \(18.0^{\circ} \mathrm{C}\) The molar mass of helium is \(4.00 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

Short Answer

Expert verified
The number of moles of helium is approximately \(0.1215 ~moles\). The pressure in the tank is roughly \(141970 ~Pa\) or \(1.4~atm\).

Step by step solution

01

Find the number of moles

First, you should convert the mass of helium to grams (since the moloar mass is given in grams/mole) and then calculate the number of moles using the formula \(n = \frac{m}{M_m}\), where \(m\) is the mass and \(M_m\) is the molar mass. Here, \(m= 4.86 \times 10^{-4} \mathrm{~kg} = 0.486 \mathrm{~g}\) and \(M_m = 4.00 \mathrm{~g/mol}\). So the number of moles \(n = \frac{0.486}{4.00} = 0.1215\) moles.
02

Convert the temperature to Kelvin

The Ideal Gas law requires the temperature to be in Kelvin. To convert from Celsius to Kelvin, you'd add 273.15 to the Celsius temperature. This makes the temperature \(T=18.0^{\circ}C+273.15 = 291.15~K\).
03

Calculate the pressure

Now you can calculate the pressure using the Ideal Gas Law. Rearranging the Law to solve for \(P\), we get \(P= \frac{nRT}{V}\). The volume \(V = 20.0~L = 0.020~m^3\), \(n = 0.1215~mol\), \(R = 8.314~J/(mol.K)\), and \(T= 291.15~K\). Substitute these values to get \(P= \frac{(0.1215)(8.314)(291.15)}{0.020}= 141970~Pa\). To convert to atmospheres, we know 1 Pa = \(9.86923 \times 10^{-6}~atm\), so the pressure in atmospheres would be \((141970)(9.86923 \times 10^{-6}) = 1.4~atm\)
04

Round the results

In order to provide the final results in an appropriate number of significant figures, the values obtained should be rounded off. Therefore, the number of moles of helium is approximately \(1.21 \times 10^{-1}~moles\), the pressure inside the tank in pascals is roughly \(1.42 \times 10^{5}~Pa\), and in atmospheres is around \(1.4~atm\).

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Key Concepts

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

Pressure Calculation
When working with gases, pressure calculations are often crucial. Using the Ideal Gas Law, we can determine the pressure within a container. This is especially true when you know the amount of gas, the volume of the tank, and the temperature. The Ideal Gas Law is expressed as:

\[ P = \frac{nRT}{V} \]
Here:
  • \( P \) is the pressure,
  • \( n \) represents the number of moles of the gas,
  • \( R \) is the universal gas constant with a value of 8.314 J/(mol路K),
  • \( T \) is the temperature in Kelvin,
  • and \( V \) is the volume in cubic meters.
It's important to always double-check your unit conversions!
For our helium tank problem, applying the given values to the formula produces a pressure of approximately 141,970 Pa. Converting this to atmospheres is simple by remembering that 1 atmosphere equals 101,325 Pa. This means approximately 1.4 atm, showing just how pressure increases with a relatively small amount of helium.
Mole Calculation
Understanding moles is a vital part of chemistry, especially when working with gases. The mole is a unit that measures the amount of substance. In our helium example, calculating the number of moles starts with converting the mass from kilograms to grams because the molar mass is given in grams/mole.

Since the mass of helium is given as \(4.86 \times 10^{-4}\) kg, which equals 0.486 g, we can then use the mole formula:

\[ n = \frac{m}{M_m} \]
Where \( n \) is the number of moles, \( m \) is the mass in grams, and \( M_m \) is the molar mass, 4.00 g/mol for helium.

Thus, \( n \) equals \( \frac{0.486}{4.00} \), amounting to about 0.1215 moles. Recognizing this small quantity of helium is crucial for calculating how it contributes to the overall gas pressure inside the tank.
Temperature Conversion
Temperature is a critical factor in the behavior of gases. In the Ideal Gas Law, we must not forget that temperature should always be represented in Kelvin. This involves converting from commonly used Celsius.

Temperature conversion from Celsius to Kelvin is straightforward:
  • Add 273.15 to the Celsius temperature to get the Kelvin figure.
For helium at 18.0掳C, the conversion is:

\[ T = 18.0 + 273.15 = 291.15~K \]
This conversion is vital because the Kelvin scale is an absolute temperature scale starting from absolute zero, where theoretically, all particle motion ceases. Using Kelvin ensures that the calculations in the Ideal Gas Law remain consistent and accurate. Take note that using Celsius directly without converting to Kelvin would lead to incorrect results.

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Most popular questions from this chapter

During a test dive in \(1939,\) prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was \(73.0 \mathrm{~m}\). The temperature was \(27.0^{\circ} \mathrm{C}\) at the surface and \(7.0^{\circ} \mathrm{C}\) at the bottom. The density of seawater is \(1030 \mathrm{~kg} / \mathrm{m}^{3}\). (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder \(2.30 \mathrm{~m}\) high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

\(\mathrm{A}\) welder using a tank of volume \(0.0750 \mathrm{~m}^{3}\) fills it with oxygen (molar mass \(32.0 \mathrm{~g} / \mathrm{mol}\) ) at a gauge pressure of \(3.00 \times 10^{5} \mathrm{~Pa}\) and temperature of \(37.0^{\circ} \mathrm{C}\). The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is \(22.0^{\circ} \mathrm{C},\) the gauge pressure of the oxygen in the tank is \(1.80 \times 10^{5} \mathrm{~Pa}\). Find (a) the initial mass of oxygen and (b) the mass of oxygen that has leaked out.

It is possible to make crystalline solids that are only one layer of atoms thick. Such "two-dimensional" crystals can be created by depositing atoms on a very flat surface. (a) If the atoms in such a twodimensional crystal can move only within the plane of the crystal, what will be its molar heat capacity near room temperature? Give your answer as a multiple of \(R\) and in \(\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\). (b) At very low temperatures, will the molar heat capacity of a two-dimensional crystal be greater than, less than, or equal to the result you found in part (a)? Explain why.

What is one reason the noble gases are preferable to air (which is mostly nitrogen and oxygen) as an insulating material? (a) Noble gases are monatomic, so no rotational modes contribute to their molar heat capacity; (b) noble gases are monatomic, so they have lower molecular masses than do nitrogen and oxygen; (c) molecular radii in noble gases are much larger than those of gases that consist of diatomic molecules; (d) because noble gases are monatomic, they have many more degrees of freedom than do diatomic molecules, and their molar heat capacity is reduced by the number of degrees of freedom.

A cylindrical diving bell has a radius of \(750 \mathrm{~cm}\) and a height of \(2.50 \mathrm{~m}\). The bell includes a top compartment that holds an undersea adventurer. A bottom compartment separated from the top by a sturdy grating holds a tank of compressed air with a valve to release air into the bell, a second valve that can release air from the bell into the sea, a third valve that regulates the entry of seawater for ballast, a pump that removes the ballast to increase buoyancy, and an electric heater that maintains a constant temperature of \(20.0^{\circ} \mathrm{C}\). The total mass of the bell and all of its apparatuses is \(4350 \mathrm{~kg}\). The density of seawater is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). (a) An \(80.0 \mathrm{~kg}\) adventurer enters the bell. How many liters of seawater should be moved into the bell so that it is neutrally buoyant? (b) By carefully regulating ballast, the bell is made to descend into the sea at a rate of \(1.0 \mathrm{~m} / \mathrm{s}\). Compressed air is released from the tank to raise the pressure in the bell to match the pressure of the seawater outside the bell. As the bell descends, at what rate should air be released through the first valve? (Hint: Derive an expression for the number of moles of air in the bell \(n\) as a function of depth \(y ;\) then differentiate this to obtain \(d n / d t\) as a function of \(d y / d t .)\) (c) If the compressed air tank is a fully loaded, specially designed, \(600 \mathrm{ft}^{3}\) tank, which means it contains that volume of air at standard temperature and pressure ( \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) ), how deep can the bell descend?
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