Chemistry Answer Note #61

Question 1 A sample of krypton gas is stored in a container with a fixed volume at 20.0 atm of pressure. The temperature (in K) of the gas is halved. What is the new pressure of the gas, assuming that the amount of gas is constant? Answer To answer this question, we need to use the ideal gas law which states that the pressure, volume, and temperature of a gas are related by the equation PV = nRT, where P is pressure in atmospheres (atm), V is volume in liters (L), n is the number of moles of gas, R is the ideal gas constant (0.0821 L*atm/mol*K), and T is temperature in Kelvin (K). Given that the amount of gas is constant, we can simplify the ideal gas law to P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the new pressure and volume. In this scenario, the initial pressure is 20.0 atm and the initial volume is constant, so we can leave V1 as it is. The temperature of the gas is halved, which means that we have a new temperature of T1/2. Therefore, we can rewrite the equation as 20.0 atm * V1 = P2 * V1 * (T1/2). We can cancel V1 from both sides, and rearrange the equation to solve for P2, which gives us P2 = (20.0 atm * T1/2)/T1. Substituting in the value of T1 in Kelvin (we need to convert from half of the temperature in K), we get P2 = (20.0 atm * 0.5T)/T = 10.0 atm. Therefore, the new pressure of the krypton gas is 10.0 atm. Question 2 A 3 L sample of a noble gas is stored in a container with a constant pressure. If the temperature (in K) of the gas is tripled, what is the new volume of the gas, assuming the container is flexible? Answer According to the Ideal Gas Law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. In this question, the pressure and number of moles are constant, therefore we can rearrange the equation to solve for the new volume. Using the rearranged equation V1/T1 = V2/T2, we can find the new volume (V2) by solving for V2 = (V1 x T2)/T1. We know that the new temperature (T2) is tripled, so T2 = 3 x T1. Therefore, the new volume (V2) is equal to V2 = (V1 x 3 x T1)/T1 = 3 x V1. In simpler terms, the new volume will be three times the initial volume. So, in this case, the new volume would be 3 L x 3 = 9 L. Question 3 Which of the following gases would have the highest density at the same temperature, pressure, and volume? Question 6 options: H2 CClF3 CO2 C2H6 CF4 Answer The gas with the highest density at the same temperature, pressure, and volume would be carbon tetrafluoride (CF4). This is because CF4 has the highest molecular weight out of the given options, making it the most dense at the same conditions. According to the ideal gas law, the density (ρ) of a gas is directly proportional to its molecular weight (M) and inversely proportional to its temperature (T), pressure (P), and molar volume (V). This concept is represented by the following equation: ρ = (PM)/(RT) Where: ρ = density (g/L) P = pressure (atm) M = molecular weight (g/mol) R = gas constant (0.0821 L·atm/mol·K) T = temperature (K) V = molar volume (L) Using this equation, we can compare the densities of the options given at the same temperature (273 K), pressure (1 atm), and molar volume (22.4 L/mol). By plugging in the values for each gas, we get the following densities: - H2: [1(2)]/[0.0821(273)] = 0.087 g/L - CClF3: [1(49.49)]/[0.0821(273)] = 2.026 g/L - CO2: [1(44.01)]/[0.0821(273)] = 1.808 g/L - C2H6: [1(30.07)]/[0.0821(273)] = 1.234 g/L - CF4: [1(88.01)]/[0.0821(273)] = 3.611 g/L From these calculations, we can see that CF4 has the highest density at the conditions given. Question 4 When a sealed 2 L flask is heated, what happens to the gas molecules? Question 4 options: They move closer together. Their kinetic energy decreases. They collide more often with the sides of the flask. They decrease their molecular motion. Their intermolecular forces increase. Answer When a sealed 2 L flask is heated, the gas molecules inside undergo various changes that can be explained through the kinetic molecular theory of gases. Before diving into the specific changes, it is essential to understand the basics of the theory. According to this theory, gases are made up of tiny particles called molecules that are constantly in motion and have negligible volume. These molecules are in continuous, random motion and possess kinetic energy. They also exert pressure on the walls of their container as a result of their constant motion. Now, coming back to the question at hand, when a sealed 2 L flask is heated, the gas molecules undergo several changes. Firstly, their kinetic energy increases as a result of the increase in temperature. This increase in kinetic energy causes the molecules to move faster, leading to an increase in their speed. This can be explained by the kinetic energy formula, where temperature is directly proportional to the average kinetic energy of the molecules (1). Secondly, the gas molecules collide more often with the sides of the flask. This is due to the increase in their speed, as explained earlier, which leads to an increase in the number of collisions per unit time. This also results in an increase in the pressure exerted by the gas on the walls of the flask (2). Additionally, the gas molecules do not move closer together in a sealed flask as there is no change in volume. The average distance between the molecules remains the same, although their speed and frequency of collisions increase. Moreover, the intermolecular forces between the gas molecules remain constant as they are not affected by changes in temperature. These forces, also known as van der Waals forces, are the attractive forces between molecules that play a role in determining the physical properties of gases (3). To further understand the changes in the gas molecules when a sealed 2 L flask is heated, let us take a look at the ideal gas law. According to this law, under constant temperature and pressure, the product of volume and pressure of a gas is directly proportional to the number of molecules (4). This means that when the flask is heated, the volume and temperature remain constant, but the number of molecules increases due to an increase in speed and frequency of collisions. However, since the flask is sealed, the volume remains the same, resulting in an increase in pressure. In conclusion, when a sealed 2 L flask is heated, the gas molecules inside exhibit an increase in kinetic energy, speed, frequency of collisions, and pressure. Their distance between molecules remains the same, and intermolecular forces do not change. This can be explained through the kinetic molecular theory of gases and the ideal gas law. Question 5 When a sealed 2 L flask is heated, what happens to the gas molecules? Answer When a sealed 2 L flask is heated, the gas molecules inside the flask will experience an increase in temperature. This increase in temperature will cause the gas molecules to have a higher average kinetic energy. As a result, the gas molecules will move at a faster rate and collide with the walls of the flask with greater force. This, in turn, will increase the pressure inside the flask. As stated by the Ideal Gas Law, PV=nRT, an increase in temperature will lead to an increase in pressure if the volume and number of moles remain constant. In this case, the volume of the flask is constant at 2 L and the number of moles of gas is also constant. Thus, the pressure inside the sealed flask will increase. Moreover, as the gas molecules collide with each other and with the walls of the flask, they will also experience an increase in speed and frequency of collisions. This will lead to an increase in the volume occupied by the gas molecules. This phenomenon is known as thermal expansion. As the gas molecules continue to gain kinetic energy and increase their speed, they may eventually reach a point where they can overcome the forces holding them together and escape from the flask. This process is known as evaporation. The rate of evaporation will also increase with an increase in temperature and can be calculated using the ClausiusClapeyron equation.