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 \(0.985\) atm at a temperature of \(-6{ }^{\circ} \mathrm{C} ;\) (b) the absolute temperature of the gas at which \(3.33 \times 10^{-3} \mathrm{~mol}\) occupies \(325 \mathrm{~mL}\) at 750 torr; \((\mathrm{c})\) the pressure, in atmospheres, if \(0.0467 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;\) (d) the quantity of gas, in moles, if \(55.7\) Lat \(54{ }^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\).

Short answer

The calculated values for each ideal gas quantity are: (a) the volume of the gas is 33.31 L, (b) the absolute temperature of the gas is 376.54 K, (c) the pressure of the gas is 4.70 atm, and (d) the quantity of gas is 2.36 mol.

Step-by-step solution

List given values and the ideal gas equation

We are given: - n = 1.50 mol - P = 0.985 atm - T = -6 °C We know the ideal gas equation is \(PV=nRT\), and we need to find the volume, V, in liters.

Convert temperature to Kelvin and use the ideal gas constant in atm

First, we need to convert the temperature from Celsius to Kelvin: K = T(°C) + 273.15 K = -6 + 273.15 K = 267.15 K Then, use the gas constant R in the appropriate unit for atm (0.0821 L·atm/mol·K).

Calculate the volume

We plug the given values into the ideal gas equation and solve for V: \(P \times V = n \times R \times T\) \(V = \frac{n \times R \times T}{P}\) \(V = \frac{1.50 \text{ mol} \times 0.0821 \frac{\text{L atm}}{\text{mol K}} \times 267.15 \text{ K}}{0.985 \text{ atm}}\) V = 33.31 L (b) Calculate the absolute temperature of the gas

List given values and the ideal gas equation

We are given: - n = \(3.33 \times 10^{-3}\) mol - V = 325 mL - P = 750 torr We know the ideal gas equation is \(PV=nRT\), and we need to find the absolute temperature T, in Kelvin.

Convert volume to liters, pressure to atm, and use the appropriate gas constant

First, we need to convert the volume from mL to L: \(325 \text{ mL} = 0.325 \text{ L}\) Next, convert pressure from torr to atm: \(750 \text{ torr} = \frac{750}{760} \text{ atm} = 0.9868 \text{ atm}\) Finally, use the gas constant R in the appropriate unit for atm (0.0821 L·atm/mol·K).

Calculate the absolute temperature

We plug the given values into the ideal gas equation and solve for T: \(P \times V = n \times R \times T\) \(T = \frac{P \times V}{n \times R}\) \(T = \frac{0.9868 \text{ atm} \times 0.325 \text{ L}}{(3.33 \times 10^{-3}) \text{ mol} \times 0.0821 \frac{\text{L atm}}{\text{mol K}}}\) T = 376.54 K (c) Calculate the pressure of the gas, in atmospheres

List given values and the ideal gas equation

We are given: - n = 0.0467 mol - V = 413 mL - T = 138 °C We know the ideal gas equation is \(PV=nRT\), and we need to find the pressure, P, in atm.

Convert volume to liters, temperature to Kelvin, and use the appropriate gas constant

First, we need to convert the volume from mL to L: \(413 \text{ mL} = 0.413 \text{ L}\) Next, convert the temperature from Celsius to Kelvin: K = T(°C) + 273.15 K = 138 + 273.15 K = 411.15 K Finally, use the gas constant R in the appropriate unit for atm (0.0821 L·atm/mol·K).

Calculate the pressure

We plug the given values into the ideal gas equation and solve for P: \(P \times V = n \times R \times T\) \(P = \frac{n \times R \times T}{V}\) \(P = \frac{0.0467 \text{ mol} \times 0.0821 \frac{\text{L atm}}{\text{mol K}} \times 411.15 \text{ K}}{0.413 \text{ L}}\) P = 4.70 atm (d) Calculate the quantity of gas, in moles

List given values and the ideal gas equation

We are given: - V = 55.7 L - T = 54 °C - P = 11.25 kPa We know the ideal gas equation is \(PV=nRT\), and we need to find the number of moles, n.

Convert pressure to atm, temperature to Kelvin, and use the appropriate gas constant

First, we need to convert the pressure from kPa to atm: \(11.25 \text{ kPa} = \frac{11.25}{101.3} \text{ atm} = 0.1111 \text{ atm}\) Next, convert the temperature from Celsius to Kelvin: K = T(°C) + 273.15 K = 54 + 273.15 K = 327.15 K Finally, use the gas constant R in the appropriate unit for atm (0.0821 L·atm/mol·K).

Calculate the quantity of gas, in moles

We plug the given values into the ideal gas equation and solve for n: \(P \times V = n \times R \times T\) \(n = \frac{P \times V}{R \times T}\) \(n = \frac{0.1111 \text{ atm} \times 55.7 \text{ L}}{0.0821 \frac{\text{L atm}}{\text{mol K}} \times 327.15 \text{ K}}\) n = 2.36 mol

Key concepts

Mole Calculations

The concept of moles is fundamental in chemistry as it allows us to count particles like atoms or molecules efficiently. A mole represents Avogadro's number, approximately 6.022 x 10^23 particles. In the ideal gas law, the mole notation 'n' signifies the number of moles of gas present in a system. When using the ideal gas equation, which is given by \(PV = nRT\), 'n' denotes the moles, R is the ideal gas constant (0.0821 L·atm/mol·K) for calculations dealing with pressure in atmospheres, volume in liters, and temperature in Kelvin. Mole calculations often involve rearranging this equation to solve for 'n', where: - To find the moles of gas, rearrange to \(n = \frac{PV}{RT}\). Understanding this principle is crucial as it allows conversion between different states of matter through stoichiometry when dealing with reactions. Practicing how to calculate moles helps solidify this basic yet essential skill in solving gases-related problems.

Pressure Conversions

Converting pressure units is necessary because the ideal gas equation uses pressure in specific units. The most common units are atmospheres (atm), torr, and kilopascals (kPa). Most exercises, including this one, require converting units like torr or kPa to atm to simplify calculations using \(PV=nRT\). - To convert torr to atm, use the conversion factor: \(1\) atm = \(760\) torr. For example, \(750\) torr is equal to \(750/760\) atm. - Kilopascals can be converted to atm using \(1\) atm = \(101.3\) kPa. For example, \(11.25\) kPa converts to \(11.25/101.3\) atm. Converting pressure correctly ensures accuracy in your results. If pressure is not converted properly, it can result in significant errors when applying the ideal gas law.

Temperature Conversions

In chemistry, temperature must often be converted from Celsius to Kelvin. Kelvin is the SI unit of temperature and is necessary for most calculations involving gases, such as those using the ideal gas law. The Kelvin scale begins at absolute zero, and its conversion from Celsius is simple: - To convert Celsius to Kelvin, add \(273.15\) to the Celsius temperature. For example, a temperature of \(-6 \degree C\) becomes \(267.15 \text{ K}\).- In collinear terms, Celsius temperatures like \(138 \degree C\) convert to \(411.15 \text{ K}\) when \(273.15\) is added. Working with Kelvin is not just a matter of preference but a necessity due to the nature of thermodynamic equations that prohibit negative values which appear on the Celsius scale.

Volume Calculations

Volume calculations are integral when using the ideal gas law, \(PV=nRT\). Volume is typically expressed in liters (L) when using this equation. For problems where volume is provided in milliliters (mL), conversion to liters is required. - Since \(1 \text{ L} = 1000 \text{ mL}\), to convert, divide the milliliter volume by \(1000\). For instance, \(325 \text{ mL}\) becomes \(0.325 \text{ L}\) after conversion.Understanding the physical significance of volume measurements can help interpret results, such as the space that the gas occupies under certain conditions. Correct volume conversion is crucial for accurate calculations, making it a foundational skill in dealing with gaseous systems.