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Chapter 18: Problem 6

You have several identical balloons. You experimentally determine that a balloon will break if its volume exceeds 0.900 L. The pressure of the gas inside the balloon equals air pressure 1.00 atm). (a) If the air inside the balloon is at a constant temperature of \(22.0^{\circ} \mathrm{C}\) and behaves as an ideal gas, what mass of air can you blow into one of the balloons before it bursts? (b) Repeat part (a) if the gas is helium rather than air.

Short Answer

Expert verified
(a) Mass of air: 1.076 g. (b) Mass of helium: 0.148 g.

Step by step solution

01

Convert Temperature

First, we need to convert the given temperature from Celsius to Kelvin since we are working with an ideal gas. The formula to convert Celsius to Kelvin is: \[ T(\text{K}) = T(\degree C) + 273.15 \]Plugging in the given temperature:\[ T = 22.0 + 273.15 = 295.15\, \text{K} \]
02

Collect Ideal Gas Law Variables

We have the following values:- Volume, \(V = 0.900\, \text{L} = 0.900 \times 10^{-3}\, \text{m}^3\) (convert liters to cubic meters)- Pressure, \(P = 1.00\, \text{atm} = 101325\, \text{Pa}\) (convert atm to pascals)- Temperature, \(T = 295.15\, \text{K}\)We will use the ideal gas law: \[ PV = nRT \]Where \(R = 8.314\, \text{J}\, \text{mol}^{-1}\, \text{K}^{-1}\) is the ideal gas constant.
03

Solve for Moles of Gas

Rearrange the ideal gas law to solve for the moles of gas, \( n \):\[ n = \frac{PV}{RT} \]Substitute the known values:\[ n = \frac{101325 \times 0.900 \times 10^{-3}}{8.314 \times 295.15} \]Calculating the above expression:\[ n \approx 0.0371\, \text{mol} \]
04

Mass of Air

We can find the mass using the molar mass of air, which is approximately \( 29.0\, \text{g/mol} \).\[ \text{Mass} = n \times \text{Molar Mass} = 0.0371 \times 29.0 = 1.076\, \text{g} \]Thus, the mass of air is approximately 1.076 gram.
05

Mass of Helium

Repeat the calculation for helium. The molar mass of helium is \( 4.0\, \text{g/mol} \):\[ \text{Mass of helium} = n \times \text{Molar Mass} = 0.0371 \times 4.0 = 0.1484\, \text{g} \]Thus, the mass of helium is approximately 0.1484 gram.

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Key Concepts

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

Balloon Burst Volume
When working with balloons and gases, knowing when a balloon will burst is important. A balloon bursts when its volume exceeds a certain limit. In this exercise, the critical volume is 0.900 liters. This means if you fill a balloon with any gas, be it air or helium, it will pop if the volume inside reaches or exceeds 0.900 liters. Understanding how much gas or air can cause a balloon to burst allows us to calculate the quantity of gas molecules needed. The gas keeps the balloon inflated, and if it expands too much from adding more gas at constant pressure, the balloon can't handle the strain and bursts. The burst volume is tied to the balloon's material and construction, but for this exercise, we only need that it can't hold more than 0.900 liters.
Temperature Conversion
When dealing with gases, especially using the Ideal Gas Law, it becomes necessary to work in Kelvin rather than Celsius. This is because Kelvin is an absolute temperature scale, which directly relates to the energy of particles.To convert temperature from Celsius to Kelvin, use the formula:
  • Convert:



Applying this to our problem, with air at 22掳C, we convert it to Kelvin:\[T = 22.0 + 273.15 = 295.15 \, \text{K}\] This conversion is crucial because it allows us to accurately apply the Ideal Gas Law in our calculations.
Molar Mass Calculation
To find out how much mass of a gas can be added to a balloon before it bursts, we need to calculate the moles of the gas and use the molar mass. The Ideal Gas Law, \(PV = nRT\), helps us determine the moles \(n\).Here's what each variable represents:
  • \(P\): Pressure of the gas (converted to pascals if originally in atm)
  • \(V\): Volume of the gas inside the balloon (preferably in cubic meters)
  • \(R\): Ideal gas constant \(8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\)
  • \(T\): Temperature of the gas in Kelvin
Rearrange the law to solve for moles:\[n = \frac{PV}{RT}\]After finding \(n\), the next step involves using the molar mass to find the mass of the gas:
  • For air, the molar mass is approximately \(29.0 \, \text{g/mol}\)
  • For helium, it's \(4.0 \, \text{g/mol}\)
Multiply the moles by the molar mass to get the mass:\[\text{Mass} = n \times \text{Molar Mass}\]So, by understanding the molar mass calculation, you can see exactly how much gas can be safely added to the balloon without exceeding the burst volume. This method helps predict and measure how different gases behave when placed under the same conditions.

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Most popular questions from this chapter

Smoke particles in the air typically have masses of the order of \(10^{-16} \mathrm{kg} .\) The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope. (a) Find the root-mean-square speed of Brownian motion for a particle with a mass of \(3.00 \times 10^{-16} \mathrm{kg}\) in air at 300 \(\mathrm{K}\) . (b) Would the root-mean-square speed be different if the particle were in hydrogen gas at the same temperature? Explain.

A cylinder 1.00 m tall with inside diameter 0.120 \(\mathrm{m}\) is used to hold propane gas (molar mass 44.1 \(\mathrm{g} / \mathrm{mol}\) ) for use in a barbecue. It is initially filled with gas until the gauge pressure is \(1.30 \times 10^{6} \mathrm{Pa}\) and the temperature is \(22.0^{\circ} \mathrm{C} .\) The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure is \(2.50 \times 10^{5}\) Pa. Calculate the mass of propane that has been used.

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 \(\mathrm{cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\) At the end of the stroke, the air has been compressed to a volume of 46.2 \(\mathrm{cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6}\) Pa. Compute the final temperature.

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 \(\mathrm{m}^{3}\) of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 \(\mathrm{m}^{3} .\) If the temperature remains constant, what is the final value of the pressure?

CP BIO The Bends. If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25 \(\mathrm{m}\) in Lake Michigan (which is fresh water), what will be the volume at the surface of an \(\mathrm{N}_{2}\) bubble that occupied 1.0 \(\mathrm{mm}^{3}\) in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due only to the changing water pressure, not to any temperature difference, an assumption that is reasonable, since we are warm-blooded creatures.)
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