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A tank having a volume of \(0.100 \mathrm{m}^{3}\) contains helium gas at 150 atm. How many balloons can the tank blow up if each filled balloon is a sphere \(0.300 \mathrm{m}\) in diameter at an absolute pressure of 1.20 atm?

Short Answer

Expert verified
Approximately 887 balloons can be inflated using the helium gas in the tank.

Step by step solution

01

Calculate the Volume of Helium at 1.20 atm

Here, we use Boyle’s law, meaning the volume is inversely proportional to pressure when temperature and quantities of gas remain constant. So, we must apply the formula: \(V_2 = V_1 * (P_1 / P_2)\). \nGiven the values: \n- Initial volume, \(V_1 = 0.100 \, m^3\) \n- Initial pressure, \( P_1 = 150 \, atm\) \n- Final pressure, \(P_2 = 1.20 \, atm\).\nUpon substituting the given values in the formula, we get: \(V_2 = 0.100 * (150 / 1.20) = 12.5 \, m^3\). This is the volume that the helium gas can fill at a pressure of 1.20 atm.
02

Calculate the Volume of a Single Balloon

The volume, \(V\), of a sphere can be calculated by the formula: \(V = (4/3)Ï€r^3\). We are given that the diameter of the balloon is 0.300 m. This means the radius, \(r = 0.150 m\).\nSubstitute the value of the radius into the volume formula to get: \(V = (4/3)Ï€ * (0.150)^3 = 0.0141 \, m^3.\)
03

Calculate the Number of Balloons

The final step is to calculate how many balloons can be filled with the given quantity of helium. That’s determined by dividing the volume of helium at 1.20 atm (from Step 1) by the volume of a single balloon (from Step 2). Therefore, the number of balloons, \(N = 12.5 / 0.0141 = 887\).

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Key Concepts

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

Boyle's Law Application
Boyle's law is an essential principle in physics that describes the inverse relationship between the pressure and the volume of a gas, at a constant temperature. In simpler terms, if you increase the pressure on a gas, its volume decreases, and conversely, if you decrease the pressure, its volume increases.

In the context of the exercise, the tank containing helium gas at 150 atm is at a much higher pressure compared to the surrounding air pressure. When we want to inflate balloons to a lesser pressure of 1.20 atm, we can apply Boyle's law to find out how much space the gas will take up at this new pressure. The formula used here translates the initial volume (\(V_1\) measured under high pressure (\(P_1\) to the new volume (\(V_2\) under lower pressure (\(P_2\) which is necessary to calculate the volume of gas we can use to inflate the balloons.

It's exact applications like this that underscore Boyle's law as foundational for understanding how gases behave under different pressures — from filling balloons to more complex tasks like operating pneumatic systems.
Helium Gas Volume Calculation
Calculating the volume of helium gas available for filling balloons involves applying the formula provided by Boyle's law, \(V_2 = V_1 * (P_1 / P_2)\). The first step is identifying the known variables from the problem: the initial volume (\(V_1\) of 0.100 m³ and the initial pressure (\(P_1\) at 150 atm. Next, we determine our desired final pressure (\(P_2\) to be 1.20 atm. Upon substituting these numbers into Boyle's law, we calculate the new volume (\(V_2\).This calculation results in a much larger volume of 12.5 m³, demonstrating how lower pressure allows the helium gas to expand. It is important to grasp how significantly gas can spread out when transitioning from high to low pressure, an insight which also applies to various real-world scenarios like weather balloons' ascension through the atmosphere where the air pressure decreases with altitude.
Balloon Volume Calculation
For a discrete object like a balloon shaped as a sphere, the volume can be determined using the formula for the volume of a sphere: \(V = (4/3)Ï€r^3\). To apply this formula, one must first obtain the sphere's radius, which is half the diameter. In the given exercise, each balloon has a diameter of 0.300 m, thus the radius is 0.150 m.

After substituting the radius into the volume formula, we obtain the volume of a single balloon. With this result, we can then figure out how many such balloons can be filled using the total available volume of helium calculated previously. It's crucial to execute each step precisely, including squaring the radius — it's a common mistake to overlook the cube in the formula. Understanding the geometry of objects is key in solving real-world problems where volumes need to be calculated, such as estimating the capacity of tanks or measuring the amount of liquid needed to fill a container.

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Most popular questions from this chapter

The mass of a hot-air balloon and its cargo (not including the air inside) is \(200 \mathrm{kg} .\) The air outside is at \(10.0^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa} .\) The volume of the balloon is \(400 \mathrm{m}^{3} .\) To what temperature must the air in the balloon be heated before the balloon will lift off? (Air density at \(10.0^{\circ} \mathrm{C}\) is \(\left.1.25 \mathrm{kg} / \mathrm{m}^{3} .\right)\)

Liquid nitrogen has a boiling point of \(-195.81^{\circ} \mathrm{C}\) at atmospheric pressure. Express this temperature (a) in degrees Fahrenheit and (b) in kelvins.

The active element of a certain laser is made of a glass rod \(30.0 \mathrm{cm}\) long by \(1.50 \mathrm{cm}\) in diameter. If the temperature of the rod increases by \(65.0^{\circ} \mathrm{C},\) what is the increase in (a) its length, (b) its diameter, and (c) its volume? Assume that the average coefficient of linear expansion of the glass is \(9.00 \times 10^{-6}\left(^{\circ} \mathrm{C}\right)^{-1}\).

(a) Use the equation of state for an ideal gas and the definition of the coefficient of volume expansion, in the form \(\beta=(1 / V) d V / d T,\) to show that the coefficient of volume expansion for an ideal gas at constant pressure is given by \(\bar{\beta}=1 / T,\) where \(T\) is the absolute temperature. (b) What value does this expression predict for \(\beta\) at \(0^{\circ} \mathrm{C} ?\) Compare this result with the experimental values for helium and air in Table \(19.1 .\) Note that these are much larger than the coefficients of volume expansion for most liquids and solids.

The average coefficient of volume expansion for carbon tetrachloride is \(5.81 \times 10^{-4}\left(^{\circ} \mathrm{C}\right)^{-1} .\) If a 50.0 -gal steel container is filled completely with carbon tetrachloride when the temperature is \(10.0^{\circ} \mathrm{C},\) how much will spill over when the temperature rises to \(30.0^{\circ} \mathrm{C} ?\)

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