Problem 43 - Chemistry The Central Science

Question

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of \(\mathrm{Cl}_{2}\) gas is 8.70 \(\mathrm{L}\) at 895 torr and \(24^{\circ} \mathrm{C}\) .(a) How many grams of \(\mathrm{Cl}_{2}\) are in the sample? (b) What volume will the \(\mathrm{Cl}_{2}\) occupy at \(\mathrm{STP}\) ? (c) At what temperature will the volume be 15.00 \(\mathrm{L}\) if the pressure is \(8.76 \times 10^{2}\) torr? (d) At what pressure will the volume equal 5.00 L if the temperature is \(58^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

(a) There are 24.9 g of \(\mathrm{Cl}_2\) in the sample. (b) The volume of the \(\mathrm{Cl}_2\) at STP is 7.83 L. (c) The temperature required for the volume to be 15.00 L at a pressure of \(8.76 \times 10^2\) torr is 601.8 K. (d) The pressure required for the volume to equal 5.00 L at a temperature of \(58^{\circ} \mathrm{C}\) is 1,467 torr.

Step 1: (a) Calculate the moles of Cl2

First, we need to convert the given pressure and temperature to the correct units for use in the ideal gas law. Convert the pressure from 895 torr to atm by dividing by 760 torr/atm. Convert the temperature from 24°C to Kelvin by adding 273.15 K. \(P = \frac{895\,\mathrm{torr}}{760\,\mathrm{torr/atm}} = 1.178\,\mathrm{atm}\) \(T = 24 + 273.15 = 297.15\,\mathrm{K}\) Next, we will use the ideal gas law, \(PV=nRT\), to calculate the number of moles of \(\mathrm{Cl}_2\). For this, we need the gas constant R in the appropriate units. We'll use \(R = 0.0821\,(\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}})\). \(n = \frac{PV}{RT} = \frac{(1.178\,\mathrm{atm})(8.70\,\mathrm{L})}{(0.0821\,(\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}}))(297.15\,\mathrm{K})}\) \(n = 0.351\,\mathrm{mol}\)

Step 2: (a) Calculate the mass of Cl2

Now that we know the number of moles of \(\mathrm{Cl}_2\), we can use the molar mass of chlorine gas to find the mass of the sample. The molar mass of Cl is 35.45 g/mol, so the molar mass of \(\mathrm{Cl}_2\) is 2 × 35.45 = 70.90 g/mol. \(m = n \times \mathrm{molar\:mass\:Cl_2} = 0.351\,\mathrm{mol} \times 70.90\,\frac{\mathrm{g}}{\mathrm{mol}} = 24.9\,\mathrm{g}\)

Step 3: (b) Calculate the volume of Cl2 at STP

To find the volume of the \(\mathrm{Cl}_2\) sample at STP (standard temperature and pressure), we need to use the ideal gas law again with P = 1 atm and T = 273.15 K. \(V_{STP} = \frac{nRT_{STP}}{P_{STP}} = \frac{(0.351\,\mathrm{mol})(0.0821\,(\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}}))(273.15\,\mathrm{K})}{1\,\mathrm{atm}}\) \(V_{STP} = 7.83\,\mathrm{L}\)

Step 4: (c) Calculate the temperature for a volume of 15.00 L at a pressure of 8.76×10² torr

Given the pressure in torr and the desired volume of 15.00 L, we need to find the temperature to achieve this volume. First, convert the pressure to atm. \(P = \frac{8.76 \times 10^2\,\mathrm{torr}}{760\,\mathrm{torr/atm}} = 1.153\,\mathrm{atm}\) Now, we can use the ideal gas law to solve for the temperature, T. \(T = \frac{PV}{nR} = \frac{(1.153\,\mathrm{atm})(15.00\,\mathrm{L})}{(0.351\,\mathrm{mol})(0.0821\,(\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}}))}\) \(T = 601.8\,\mathrm{K}\)

Step 5: (d) Calculate the pressure for a volume of 5.00 L at a temperature of 58°C

Lastly, we need to find the pressure to achieve a volume of 5.00 L at a given temperature of 58°C. Convert the temperature to Kelvin. \(T = 58 + 273.15 = 331.15\,\mathrm{K}\) Now, we can use the ideal gas law to solve for the pressure, P. \(P = \frac{nRT}{V} = \frac{(0.351\,\mathrm{mol})(0.0821\,(\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}}))(331.15\,\mathrm{K})}{5.00\,\mathrm{L}}\) \(P = 1.931\,\mathrm{atm}\) Convert the pressure back to torr: \(P = 1.931\,\mathrm{atm} \times 760\,\mathrm{torr/atm} = 1,467\,\mathrm{torr}\)

Key Concepts

Molar Mass

Understanding the molar mass is crucial in chemistry, especially when it comes to ideal gas law calculations. Molar mass refers to the mass of one mole of a given substance and is measured in grams per mole (g/mol). To find the molar mass of a molecular compound such as chlorine gas (\r \(\mathrm{Cl}_{2}\)), we sum the molar masses of its constituent atoms. Since chlorine has an atomic molar mass of about 35.45 g/mol, the molar mass of \r \(\mathrm{Cl}_{2}\) is simply twice that, resulting in 70.90 g/mol.

When dealing with gas law problems, knowing the molar mass allows us to transition between the number of moles and the mass of a gas. For example, if we calculate the number of moles of chlorine gas using the ideal gas law, we can then multiply the moles by the molar mass to find the total mass of the chlorine in the sample. This is particularly handy when one needs to measure out a specific quantity of a gas for reactions or when determining the mass based on experimental gas volumes.

Standard Temperature and Pressure (STP)

When working with gases, it's often useful to reference a standard set of conditions. Standard Temperature and Pressure (STP) is a common reference point in chemistry which corresponds to a temperature of 0 degrees Celsius (273.15 K) and a pressure of 1 atmosphere (atm). At STP, one mole of an ideal gas occupies 22.4 liters, which becomes a useful conversion factor for solving gas problems.

Comparing gas volumes at different conditions is simplified by using STP as a baseline. For instance, if you know the volume of a gas at STP, you can predict its volume at other temperatures and pressures using the ideal gas law. Conversely, you can calculate what the volume would be at STP if given the volume at non-standard conditions, as demonstrated in the textbook solution for finding the volume of chlorine gas at STP.

Gas Constant R

The gas constant, denoted as R, is a fundamental aspect of the ideal gas law, which itself is an equation of state for an ideal gas. R serves as a proportionality constant that relates the pressure, volume, temperature, and mole quantity of a gas. It's essential to use the appropriate value of R that matches the units of pressure, volume, and temperature in your calculations.

The most commonly used value of R is 0.0821 \r \(\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}}\) because it conveniently corresponds with the pressure in atmospheres, volume in liters, and temperature in Kelvin, which are typical units for gas law problems. Employing the correct value of R is crucial in achieving accurate results when using the ideal gas law.

Converting Units in Gas Law Problems

In the context of the ideal gas law and associated calculations, it's critical to ensure all your units are compatible. You'll often need to convert between various units of pressure (like atmospheres, torr, and Pascal), volume (liters, milliliters, cubic meters), and temperature (Celsius to Kelvin).

Remember, for pressure, 1 atm is equivalent to 760 torr. Temperature must always be in Kelvin in gas law equations, which can be converted from Celsius by adding 273.15. Volume conversions are more direct; 1000 milliliters make up 1 liter, for example. These conversions matter because using mismatched or incorrect units can throw off the entire calculation, leading to erroneous results. Therefore, always convert your units before plugging values into your ideal gas law equation.