Problem 34 - Chemistry The Central Science In Si Units

Question

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 \(11.25 \mathrm{kPa} .\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \(126.7\, \mathrm{kPa} = 126700 \, \mathrm{Pa}\).

Step 3: 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, \( V \): \[ V = \frac{nRT}{P} = \frac{1.50 \times 8.314 \times 267.15}{126700} \approx 0.0263 \mathrm{~m^3} = 26.3 \mathrm{~L} \]

Step 4: Convert Units for (b) and Solve for Temperature

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

Step 5: 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 \( P \): \[ P = \frac{nRT}{V} = \frac{0.00245 \times 8.314 \times 411.15}{0.413} \approx 20,330 \mathrm{~Pa}. \]

Step 6: 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 \( PV = nRT \): \[ n = \frac{PV}{RT} = \frac{11250 \times 0.1265}{8.314 \times 327.15} \approx 0.0527 \mathrm{~mol}. \]

Key Concepts

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 (m³),
• \( R \) is the Ideal Gas Constant, which is 8.314 J/mol·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 (m³) 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 m³ = 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.