Question

Calculate the pressure required to compress 2 liters of a gas at \(700 \mathrm{~mm}\) pressure and \(20^{\circ} \mathrm{C}\) into a container of \(0.1\) liter capacity at a temperature of \(-150^{\circ} \mathrm{C}\).

Short answer

To calculate the final pressure required to compress the given gas, we use the Ideal Gas Law and the formula \( P_2 = P_1V_1T_2/(V_2T_1) \). After converting the temperatures to Kelvin and the initial pressure to atmospheres, we find that the final pressure \( P_2 = (700/760)(2)(123.15)/ (0.1)(293.15) \) atm.

Step-by-step solution

Convert temperature to Kelvin

Convert the initial and final temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is \[ T(K) = T(C) + 273.15 \] The initial temperature \( T_1 \) is then 20 + 273.15 = 293.15K, and the final temperature \( T_2 \) is -150 + 273.15 = 123.15K.

Convert pressure to the same units

Convert the initial pressure from mm to atm. The conversion factor is 1 atm = 760 mm Hg. The initial pressure \( P_1 \) is then 700/760 atm.

Using the Ideal Gas Law for initial and final states

Set up the Ideal Gas Law for the initial and final states of the gas, and since the number of moles is the same, the \( nRT \) term will be the same for both states. The equation then becomes \( P_1V_1/T_1 = P_2V_2/T_2 \), where \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume and temperature, and \( P_2 \), \( V_2 \), and \( T_2 \) are the final pressure, volume and temperature.

Solve for the final pressure

Substitute the given values and solve for the final pressure \( P_2 \). The equation becomes \( P_2 = P_1V_1T_2/(V_2T_1) \). Substituting the known values gives \( P_2 = (700/760)(2)(123.15)/ (0.1)(293.15) \) atm. This gives us the final pressure in at atmospheres. You can convert this value to any other units if required.

Key concepts

Gas Compression

Gas compression is a process where the volume of a gas is reduced while conserving the amount of gas molecules, which therefore increases its pressure. This is often achieved by pushing the gas into a smaller container. An example from everyday life could be inflating a bicycle tire; the air compressor forces more air into the fixed volume of the tire, thus increasing the air pressure inside.

In our exercise, we're compressing gas from a volume of 2 liters into just 0.1 liters. From the Ideal Gas Law equation \( PV = nRT \), where \( n \) remains constant for a given amount of gas, reducing \( V \) necessitates an increase in \( P \) if \( T \) also decreases. This relationship showcases how interconnected these variables are and how changes in one will affect the others when dealing with gas laws.

The process of gas compression is crucial in various applications, such as refrigeration, internal combustion engines, and industrial gas transportation. Understanding this principle aids students not only in solving physical chemistry problems but also in comprehending the mechanisms behind technologies that harness the behavior of gases under different conditions.

Temperature Conversion

Temperature conversion is fundamental in chemistry calculations, especially when using the Ideal Gas Law, which requires temperature to be measured in Kelvin. The Kelvin scale is an absolute temperature scale starting at zero, or absolute zero, where theoretically there is no particle motion.

Converting Celsius to Kelvin, as shown in the exercise, is done by adding 273.15 to the Celsius temperature. This conversion ensures that temperature values are consistent with the absolute scale used in gas law equations. It's important to convert temperatures carefully, as errors in this step can result in significant mistakes in the final answer.

To convert from Celsius to Kelvin:

• For positive Celsius temperatures: add 273.15.
• For negative Celsius temperatures: add 273.15 as well.

So, for a temperature of \( 20^\circ C \), we get 293.15 K, and for \( -150^\circ C \) we get 123.15 K. In chemistry, correct temperature conversion is the backbone to accurately determine the effects of temperature on gas pressure and volume.

Pressure Conversion

Pressure conversion is a crucial step in gas law problems because pressure can be measured using several different units, such as atmospheres (atm), millimeters of mercury (mmHg), or Pascals (Pa). In general chemistry, standard units of pressure are typically atmospheres or Pascals, but medical and other applications often use mmHg.

For our exercise, we need to convert the given pressure in mmHg into atmospheres, so we can use it in the Ideal Gas Law. The conversion factor is 1 atm = 760 mmHg. Therefore, to convert the initial pressure of 700 mmHg to atm, you would perform the following calculation: \( P_{\text{atm}} = \frac{700 \text{ mmHg}}{760 \text{ mmHg/atm}} \) which simplifies to approximately 0.9211 atm.

Understanding how to convert pressure units is fundamental in various scientific studies, including atmospheric science, engineering, and of course, chemistry. Accurate pressure conversions ensure compatibility between different scientific data sets and can make or break an experiment or industrial process.

General Chemistry

In general chemistry, the Ideal Gas Law is one of the most important equations that illustrates the relationship between pressure (P), volume (V), temperature (T), and the amount of gas in moles (n) with the universal gas constant (R). The law is commonly written as \( PV = nRT \). This law is a combination of Boyle's Law, Charles's Law, Avogadro's Law, and Gay-Lussac's Law, which individually describe the relationship between two variables while holding others constant.

In practical applications, the Ideal Gas Law allows for calculations regarding the behavior of gases under various conditions. For instance, when you need to estimate how much pressure is needed to compress a certain volume of gas at one temperature into a smaller volume at a different temperature, the Ideal Gas Law provides the toolset for prediction and understanding of these conditions.

Furthermore, the law also underpins the understanding of thermodynamics, reaction kinetics, and even real-world industrial processes that involve gas-phase reactions or transport. By engaging with these concepts, students develop problem-solving skills that are widely applicable across scientific disciplines and industries, truly embodying the integrative nature of general chemistry.