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Chapter 10: Problem 34

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(126.7 \mathrm{kPa}\) at a temperature of \(-6^{\circ} \mathrm{C} ;(\mathbf{b})\) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at \(99.99 \mathrm{kPa} ;(\mathbf{c})\) the pressure, in pascals, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;\) (d) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa} .\)

Short Answer

Expert verified
(a) 26.3 L; (b) 1728 K; (c) 20,330 Pa; (d) 0.0527 mol.

Step by step solution

01

Identify the Ideal Gas Law Formula

The ideal gas law formula is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 \( \mathrm{J \cdot mol^{-1} \cdot K^{-1}} \)), and \( T \) is temperature in Kelvin. This formula will be used for all parts of the problem.
02

Convert Units for (a)

Convert the temperature from Celsius to Kelvin by adding 273.15: \(-6^{\circ} \mathrm{C} = 267.15 \mathrm{~K}\). The pressure is already in kilopascals, appropriate for our calculations as \(126.7\, \mathrm{kPa} = 126700 \, \mathrm{Pa}\).
03

Solve for Volume in Part (a)

Using the ideal gas law \( PV = nRT \), substitute the known quantities: \( P = 126700 \mathrm{~Pa}, \, n = 1.50 \mathrm{~mol}, \) \( T = 267.15 \mathrm{~K}\). Solve for volume, \( V \): \[ V = \frac{nRT}{P} = \frac{1.50 \times 8.314 \times 267.15}{126700} \approx 0.0263 \mathrm{~m^3} = 26.3 \mathrm{~L} \]
04

Convert Units for (b) and Solve for Temperature

Convert volume from milliliters to liters: \( 478 \mathrm{~mL} = 0.478 \mathrm{L}. \) Using \( PV = nRT \), \[ T = \frac{PV}{nR} = \frac{99.99 \times 10^3 \times 0.478}{3.33 \times 10^{-3} \times 8.314} \approx 1728 \mathrm{~K}. \]
05

Convert Units for (c) and Solve for Pressure

Convert volume from milliliters to liters: \( 413 \mathrm{~mL} = 0.413 \mathrm{~L}.\) Convert temperature to Kelvin: \( 138^{\circ} \mathrm{C} = 411.15 \mathrm{~K}.\) Using \( PV = nRT \), solve for \( P \): \[ P = \frac{nRT}{V} = \frac{0.00245 \times 8.314 \times 411.15}{0.413} \approx 20,330 \mathrm{~Pa}. \]
06

Convert Units for (d) and Solve for Moles

Convert volume to cubic meters: \( 126.5 \mathrm{~L} = 0.1265 \mathrm{~m^3}. \) Convert temperature to Kelvin: \( 54^{\circ} \mathrm{C} = 327.15 \mathrm{~K}. \) Convert pressure to pascals: \( 11.25 \, \mathrm{kPa} = 11250 \, \mathrm{Pa}.\) Solve for \( n \) using \( PV = nRT \): \[ n = \frac{PV}{RT} = \frac{11250 \times 0.1265}{8.314 \times 327.15} \approx 0.0527 \mathrm{~mol}. \]

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Key Concepts

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

Mole Calculations
When dealing with gases in chemistry, it is often necessary to calculate the number of moles present. Moles are an essential concept because they provide a link between the macroscopic world of grams and liters, and the microscopic world of molecules and atoms.

To calculate moles using the Ideal Gas Law, use the formula: \( n = \frac{PV}{RT} \). Here:
  • \( P \) is Pressure measured in pascals (Pa),
  • \( V \) is Volume in cubic meters (m鲁),
  • \( R \) is the Ideal Gas Constant, which is 8.314 J/mol路K, and
  • \( T \) is Temperature in Kelvin (K).
By rearranging the Ideal Gas Law formula \( PV = nRT \) to solve for \( n \), the quantity of gas in moles can be determined easily given any three of the other variables. Understanding this calculation is key when you want to determine how much gas is present in a given volume under certain conditions of temperature and pressure.
Temperature Conversion
Temperature conversion is an indispensable skill when working with gases. In most gas calculations, temperature should be in Kelvin because it is the only absolute temperature scale used in these equations. A simple way to convert Celsius to Kelvin is by adding 273.15 to the Celsius temperature.

For example: converting \(-6^{\circ} \, \text{C}\) to Kelvin is done as follows:
  • \(-6^{\circ} \, \text{C} + 273.15 = 267.15 \, \text{K}\)
Using Kelvin avoids any chance of calculations yielding negative temperatures, which could occur in Celsius with temperatures below freezing. This conversion ensures that calculations conform to the Ideal Gas Law equations, making them consistent and reliable in understanding gas behavior.
Pressure Conversion
Pressure is another critical factor in gas calculations. It's often necessary to convert pressure values into proper units so that they can be used with the Ideal Gas Law. The standard unit used in gas calculations is the pascal (Pa).

To convert from kilopascals (kPa) to pascals, simply multiply by 1,000, since 1 kPa = 1,000 Pa. For example:
  • **126.7 kPa = 126,700 Pa**
This conversion is crucial as it ensures the values are compatible with the units of the ideal gas constant \( R \), which is expressed in joules (J). Accurate conversions lead to valid results in gas law problems.
Volume Conversion
Volume conversion is critical when working with gases, as the Ideal Gas Law often requires volume to be in specific units. Commonly, volumes need to be in liters (L) or cubic meters (m鲁) since liters are the most practical unit for typical lab settings, while cubic meters are necessary for consistency with the units of the gas constant \( R \).

For smaller volumes given in milliliters (mL), convert by:
  • 1 L = 1,000 mL
  • To convert mL to L, divide by 1,000.For example, a volume of 478 mL becomes:
  • \(478 \, \text{mL} = 0.478 \, \text{L}\)
Alternatively, converting liters to cubic meters involves dividing by 1,000 as well, since 1 m鲁 = 1,000 L. Volume conversions need to be correctly executed to ensure accuracy in calculating the behavior of gases using the Ideal Gas Law formulas.

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Most popular questions from this chapter

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is \(8.102 \mathrm{~g}\). (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. \((\mathbf{e})\) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s) $$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(152.0 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of \(2 ?\) [Section 10.3\(]\) (a) It would contain the same number of molecules. (b) It would contain half as many molecules. (c) It would contain twice as many molecules. (d) There is insufficient data to say.

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; (b) the temperature of the gas is increased at constant volume?

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(304 \mathrm{~Pa}\) and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level ( \(101.33 \mathrm{kPa}\) ) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?
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