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 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 \(6.00 \mathrm{~L}\) if the temperature is \(58^{\circ} \mathrm{C}\) ?

Short answer

(a) The sample contains 24.3 grams of \(\mathrm{Cl}_{2}\) gas. (b) The volume of \(\mathrm{Cl}_{2}\) gas at STP is approximately 7.66 L. (c) The temperature when volume is \(15.00 \mathrm{~L}\) and pressure is \(8.76\times10^2 \mathrm{~torr}\) is approximately 576.44 K. (d) The pressure when volume is \(6.00 \mathrm{~L}\) and temperature is \(58^{\circ}\mathrm{C}\) is approximately 1.969 atm.

Step-by-step solution

(a) Finding the grams of \(\mathrm{Cl}_{2}\) gas

First, let's find the number of moles (n) of the gas using the Ideal Gas Law equation: \[n = \frac{PV}{RT}\] Given values: P (pressure) = 895 torr (we need to convert it to atm: 1 atm = 760 torr) V (volume) = 8.70 L R (Ideal Gas Constant) = 0.0821 \(\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}}\) T (temperature) = \(24^{\circ}\mathrm{C}\) (convert to Kelvin by adding 273.15) Converting the pressure and temperature to appropriate units: \[P = \frac{895\ \mathrm{torr}}{760\ \mathrm{torr/atm}} = 1.178 \mathrm{~atm}\] \[T = 24^{\circ}\mathrm{C} + 273.15 = 297.15 \mathrm{~K}\] Plugging these values into the equation, we get: \[n = \frac{(1.178 \mathrm{~atm})(8.70 \mathrm{~L})}{(0.0821 \frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{mol}\cdot \mathrm{K}})(297.15 \mathrm{~K})} = 0.342 \mathrm{~mol}\] Now, we will use the molar mass of \(\mathrm{Cl}_{2}\) to convert moles to grams. The molar mass of \(\mathrm{Cl}_{2}\) is 70.90 g/mol. Mass of the \(\mathrm{Cl}_{2}\) gas sample: \[m = n \times \text{Molar Mass} = 0.342 \mathrm{~mol} \times 70.90 \frac{\mathrm{g}}{\mathrm{mol}} = 24.3 \mathrm{~g}\] The sample contains 24.3 grams of \(\mathrm{Cl}_{2}\) gas.

(b) Finding the volume of \(\mathrm{Cl}_{2}\) gas at STP

To find the volume at STP (Standard Temperature and Pressure: \(0^{\circ}\mathrm{C}\) and 1 atm), we can use the combined gas law, which is: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] Where we already know P1, V1, and T1 from part (a), and we are looking for V2. P2 and T2 are the Standard Temperature and Pressure: T2 = 0°C (convert to Kelvin by adding 273.15) P2 = 1 atm Converting temperature to Kelvin: \[T_2 = 0^{\circ}\mathrm{C} + 273.15 = 273.15 \mathrm{~K}\] Plug in the values into the combined gas law equation \[\frac{(1.178 \mathrm{~atm})(8.70 \mathrm{~L})}{297.15 \mathrm{~K}} = \frac{(1 \mathrm{~atm})(V_2)}{273.15 \mathrm{~K}}\] Now solve for \(V_2\): \[V_2 = \frac{(1 \mathrm{~atm})(8.70 \mathrm{~L})(273.15 \mathrm{~K})}{(1.178 \mathrm{~atm})(297.15 \mathrm{~K})} = 7.66 \mathrm{~L}\] The volume of \(\mathrm{Cl}_{2}\) gas at STP is approximately 7.66 L.

(c) Finding the temperature when volume is \(15.00 \mathrm{~L}\) and pressure is \(8.76\times10^2\) torr

We are given a new volume and pressure, and have to find the new temperature by using the combined gas law equation: \[\frac{P_1V_1}{T_1} = \frac{P_3V_3}{T_3}\] Given values: V3 = 15.00 L P3 = \(8.76\times10^2\) torr (convert to atm: 1 atm = 760 torr) Converting the pressure to appropriate units: \[P_3 = \frac{8.76\times10^2 \mathrm{~torr}}{760\ \mathrm{torr/atm}} \approx 1.152 \mathrm{~atm}\] Plug in the values into the combined gas law equation: \[\frac{(1.178 \mathrm{~atm})(8.70 \mathrm{~L})}{297.15 \mathrm{~K}} = \frac{(1.152 \mathrm{~atm})(15.00 \mathrm{~L})}{T_3}\] Now solve for \(T_3\): \[T_3 = \frac{(1.152 \mathrm{~atm})(15.00 \mathrm{~L})(297.15 \mathrm{~K})}{(1.178 \mathrm{~atm})(8.70 \mathrm{~L})} \approx 576.44 \mathrm{~K}\] The temperature when volume is \(15.00 \mathrm{~L}\) and pressure is \(8.76\times10^2 \mathrm{~torr}\) is approximately 576.44 K.

(d) Finding the pressure when volume is \(6.00 \mathrm{~L}\) and temperature is \(58^{\circ}\mathrm{C}\)

We are given a new volume and temperature, and have to find the new pressure by using the combined gas law equation: \[\frac{P_1V_1}{T_1} = \frac{P_4V_4}{T_4}\] Given values: V4 = 6.00 L T4 = \(58^{\circ}\mathrm{C}\) (convert to Kelvin by adding 273.15) Converting the temperature to Kelvin: \[T_4 = 58^{\circ}\mathrm{C} + 273.15 = 331.15 \mathrm{~K}\] Plug in the values into the combined gas law equation: \[\frac{(1.178 \mathrm{~atm})(8.70 \mathrm{~L})}{297.15 \mathrm{~K}} = \frac{P_4(6.00 \mathrm{~L})}{331.15 \mathrm{~K}}\] Now solve for \(P_4\): \[P_4 = \frac{(1.178 \mathrm{~atm})(8.70 \mathrm{~L})(331.15 \mathrm{~K})}{(6.00 \mathrm{~L})(297.15 \mathrm{~K})} \approx 1.969 \mathrm{~atm}\] The pressure when volume is \(6.00 \mathrm{~L}\) and temperature is \(58^{\circ}\mathrm{C}\) is approximately 1.969 atm.

Key concepts

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure, or STP, are specific reference conditions that are extremely useful when studying gases and their behaviors. STP is set at a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (atm). At STP, one mole of an ideal gas occupies a volume of 22.4 liters; this is a consequence of the ideal gas law.

The use of STP allows chemists and physicists to easily compare the behavior of different gases and predict how they will react under these standard conditions. When dealing with gas law problems, it is often necessary to convert given conditions to or from STP to apply the relevant equations correctly. For instance, in the exercise provided, the volume of chlorine gas at STP was calculated using the combined gas law by comparing the original and standard conditions.

Combined Gas Law

The Combined Gas Law is a fusion of three fundamental laws related to gases: Boyle's Law, Charles's Law, and Gay-Lussac's Law. It relates pressure, volume, and temperature of a fixed amount of gas.

The formula for the combined gas law is: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) where \(P_1\), \(V_1\), and \(T_1\) are the initial conditions, and \(P_2\), \(V_2\), and \(T_2\) are the new conditions after a change has occurred.

In our exercise, this law enables us to solve parts (b), (c), and (d), as it can be manipulated to solve for the unknown variable when the conditions of a gas sample change. It's essential to remember that temperatures must be expressed in Kelvin when using any gas laws to ensure the correct math behind the relationships.

Molar Mass of Gases

The molar mass of a gas is the mass of one mole of its molecules, typically expressed in grams per mole (g/mol). The molar mass is a critical component when calculating the amount of gas present in a sample, as demonstrated in the problem for finding the mass of chlorine gas.

The ideal gas law equation () can be used to find the number of moles of gas when rearranged as: \(m = n \times \text{Molar Mass}\). Once the number of moles () is determined, the mass () can be calculated if the molar mass is known. The molar mass of chlorine gas () used in the exercise is 70.90 g/mol, which allows us to convert moles to grams to find the mass of the gas sample.

It's important to recognize the molar mass as a conversion factor between the mass of a substance and the amount in moles, providing a bridge between the microscopic scale of atoms and molecules and the macroscopic world of grams and liters.