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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 22.0\(^\circ\)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) 1.072 g of air, (b) 0.148 g of helium.

Step by step solution

01

Understand the Ideal Gas Law

The Ideal Gas Law states that \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L路atm/mol路K), and \( T \) is the temperature in Kelvin. We will use this equation to find the number of moles of gas in the balloon.
02

Convert Temperature to Kelvin

First, convert the temperature from Celsius to Kelvin with the formula \( T(K) = T(^\circ C) + 273.15 \). The given temperature is 22.0\(^\circ\)C. Thus, \( T(K) = 22.0 + 273.15 = 295.15 \) K.
03

Calculate Moles of Air Using Ideal Gas Law

Use the Ideal Gas Law to find the number of moles of air that can be blown into the balloon. Given \( P = 1.00 \) atm, \( V = 0.900 \) L, and \( T = 295.15 \) K, solve for \( n \) in \( PV = nRT \): \( n = \frac{PV}{RT} = \frac{(1.00 \text{ atm}) (0.900 \text{ L})}{(0.0821 \text{ L路atm/mol路K})(295.15 \text{ K})} \approx 0.037 \) moles.
04

Convert Moles of Air to Mass

Find the molar mass of air, which is approximately 28.97 g/mol. Compute the mass using the moles from Step 3: \( ext{mass} = n \times ext{molar mass} = 0.037 \text{ mol} \times 28.97 \text{ g/mol} \approx 1.072 \text{ g} \).
05

Calculate Moles and Mass of Helium

For helium, use the same number of moles calculated in Step 3. The molar mass of helium is 4.00 g/mol. Compute the mass: \( ext{mass} = 0.037 \text{ mol} \times 4.00 \text{ g/mol} \approx 0.148 \text{ g} \).

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Key Concepts

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

Mole Calculation
When working with gases and the Ideal Gas Law, calculating moles is a central step. Moles are a unit used to express the amount of a substance. The Ideal Gas Law equation, \( PV = nRT \), comes in handy to find the number of moles, \( n \). Here, \( P \) stands for pressure, \( V \) is volume, \( R \) is the ideal gas constant (0.0821 L路atm/mol路K), and \( T \) is temperature in Kelvin. To use this equation:
  • Ensure all the parameters are in the correct units: Pressure in atmospheres, Volume in liters, Temperature in Kelvin.
  • Rearrange the formula to solve for the number of moles: \( n = \frac{PV}{RT} \).
For example, given the values in the exercise:
  • Pressure (\( P \)) = 1.00 atm
  • Volume (\( V \)) = 0.900 L
  • Temperature (\( T \)) = 295.15 K
You can compute the moles of gas in the balloon: \( n = \frac{(1.00 \text{ atm}) \times (0.900 \text{ L})}{(0.0821 \text{ L路atm/mol路K}) \times (295.15 \text{ K})} \approx 0.037 \text{ moles} \). This tells us how much gas (in moles) the balloon can hold before it pops.
Temperature Conversion
In gas calculations, it is crucial to work with the correct temperature units. The Ideal Gas Law demands temperature in Kelvin rather than Celsius. Kelvin is the absolute temperature scale, which doesn't have negative values and is necessary for accurate gas calculations. To convert Celsius to Kelvin:
  • Use the conversion formula: \( T(K) = T(^\circ C) + 273.15 \).
For instance, in the exercise, the given temperature is 22.0掳C. Converting this to Kelvin:
  • \( T(K) = 22.0 + 273.15 = 295.15 \text{ K} \).
Simple calculations like this ensure precision in your gas laws applications. Always remember, working in Kelvin is a must for using the Ideal Gas Law, as it ensures proportional and non-negative temperature readings.
Molar Mass
Understanding molar mass is key when converting moles into grams. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). This concept helps us to convert between the mass of a substance and the number of moles, making it practically applicable in calculations.For air, the molar mass is approximately 28.97 g/mol. We use this value to find how much mass the computed moles translate into. You can find the mass from moles using the formula:
  • \( \text{mass} = n \times \text{molar mass} \).
Similarly, for helium:
  • The molar mass is 4.00 g/mol.
So, if you have 0.037 moles of a gas like air:
  • \( \text{mass} = 0.037 \text{ mol} \times 28.97 \text{ g/mol} \approx 1.072 \text{ g} \).
And for helium:
  • \( \text{mass} = 0.037 \text{ mol} \times 4.00 \text{ g/mol} \approx 0.148 \text{ g} \).
This approach converts theoretical gas volume into tangible, real-world mass metrics, providing a deeper understanding of chemical quantities.

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Most popular questions from this chapter

A vertical cylindrical tank contains 1.80 mol of an ideal gas under a pressure of 0.300 atm at 20.0\(^\circ\)C. The round part of the tank has a radius of 10.0 cm, and the gas is supporting a piston that can move up and down in the cylinder without friction. There is a vacuum above the piston. (a) What is the mass of this piston? (b) How tall is the column of gas that is supporting the piston?

Modern vacuum pumps make it easy to attain pressures of the order of 10\({^-}{^1}{^3}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of 9.00\(\times\) 10\({^-}{^1}{^4}\) atm and an ordinary temperature of 300.0 K, how many molecules are present in a volume of 1.00 cm\(^3\)? (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

At an altitude of 11,000 m (a typical cruising altitude for a jet airliner), the air temperature is -56.5\(^\circ\)C and the air density is 0.364 kg/m\(^3\). What is the pressure of the atmosphere at that altitude? (Note: The temperature at this altitude is not the same as at the surface of the earth, so the calculation of Example 18.4 in Section 18.1 doesn't apply.)

Estimate the number of atoms in the body of a 50-kg physics student. Note that the human body is mostly water, which has molar mass 18.0 g/mol, and that each water molecule contains three atoms.

A balloon of volume 750 m\(^3\) is to be filled with hydrogen at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa). (a) If the hydrogen is stored in cylinders with volumes of 1.90 m\(^3\) at a gauge pressure of 1.20 \(\times\) 10\(^6\) Pa, how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if both the gas in the balloon and the surrounding air are at 15.0\(^\circ\)C? The molar mass of hydrogen (H\(_2\)) is 2.02 g/mol. The density of air at 15.0\(^\circ\)C and atmospheric pressure is 1.23 kg/m\(^3\). See Chapter 12 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 g/mol) instead of hydrogen, again at 15.0\(^\circ\)C?
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