Question

A sample of an ideal gas at \(15.0 \mathrm{~atm}\) and \(10.0 \mathrm{~L}\) is allowed to expand against a constant external pressure of \(2.00 \mathrm{~atm}\) at a constant temperature. Calculate the work in units of \(\mathrm{kJ}\) for the gas expansion. (Hint: Boyle's law applies.)

Short answer

The work done by the gas during expansion is approximately \(-13.17 \mathrm{~kJ}\).

Step-by-step solution

1. Applying Boyle's Law

: Boyle's Law states that for a given amount of gas at constant temperature, the pressure of the gas multiplied by its volume is constant. Mathematically, this can be expressed as: \[P_1V_1 = P_2V_2\] where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume respectively. We are given the initial pressure and volume, and the final pressure (which is equal to the external pressure). We can use Boyle's Law to find the final volume.

2. Finding the final volume

: From Boyle's Law, \(P_1V_1 = P_2V_2\). Given, \(P_1 = 15.0 \mathrm{~atm}\), \(V_1 = 10.0 \mathrm{~L}\), and \(P_2 = 2.00 \mathrm{~atm}\). So, we can find the final volume as: \[V_2 = \frac{P_1V_1}{P_2} = \frac{(15.0 \mathrm{~atm})(10.0 \mathrm{~L})}{2.00 \mathrm{~atm}} = 75.0 \mathrm{~L}\]

3. Calculating work done by the gas

: We can calculate the work done by the gas during expansion as: \[W = -P_{ext}(V_2 - V_1)\] where \(W\) is the work done, \(P_{ext}\) is the external pressure, and \(V_1\) and \(V_2\) are the initial and final volumes. Given \(P_{ext} = 2.00 \mathrm{~atm}\), \(V_1 = 10.0 \mathrm{~L}\), and \(V_2 = 75.0 \mathrm{~L}\), we can plug in the values to get: \[W = -(2.00 \mathrm{~atm})(75.0 \mathrm{~L} - 10.0 \mathrm{~L})\]

4. Converting atm to J

: Before calculating the work done, we need to convert the pressure from atm to J. The conversion factor is 1 atm = 101.325 J/L. Thus: \[P_{ext} = 2.00 \mathrm{~atm} \times \frac{101.325 \mathrm{~J}}{1 \mathrm{~L}} = 202.65 \mathrm{~atm.L^{-1}}\]

5. Calculating the work done

: Now that we have the pressure in J/L, we can plug in the values into the work formula: \[W = -(202.65 \mathrm{~J/L})(65.0 \mathrm{~L}) = -13172.25 \mathrm{~J}\]

6. Converting the work done to kJ

: To convert the work done in J to kJ, we simply divide by 1000: \[W = -13.17 \mathrm{~kJ}\] The work done by the gas during expansion is approximately -13.17 kJ.

Key concepts

Gas Expansion

When considering gas expansion, imagine a gas pushing against an external pressure that is lower than the gas's initial pressure. In this scenario, the gas expands and increases its volume. The easier it is for a gas to expand, the more work it can do. This is often seen in processes where gases exert force as they expand within a certain volume or space.
The expansion of a gas is fundamental to understanding how engines and cooling systems operate, as well as natural phenomena like the inflation of a hot air balloon. Ensuring that the temperature remains constant during expansion is crucial because varying temperatures can affect the pressure and volume in unpredictable ways.

Ideal Gas Law

The Ideal Gas Law is a cornerstone of understanding the behavior of gases under different conditions. This law provides a clear relationship between pressure, volume, and temperature in an ideal situation. While Boyle's Law, which is a subset of the Ideal Gas Law, involves constant temperature, the Ideal Gas Law extends to include temperature as a variable.
The formula for the Ideal Gas Law is: \[PV = nRT\]Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. Though not directly used in our calculation due to constant temperature, the Ideal Gas Law is critical for understanding how gas behavior changes with different variables.

Work Calculation

Calculating work, particularly in the context of gas expansion, involves understanding how much energy is transferred as the gas expands against an external pressure. The formula used to determine the work done by the gas is:\[W = -P_{ext}(V_2 - V_1)\]where \(W\) is the work done, \(P_{ext}\) is the constant external pressure, and \(V_1\) and \(V_2\) are the initial and final volumes of the gas, respectively.
Work is calculated in units of energy (joules or kilojoules). The negative sign indicates that the system (gas) is doing work on the surroundings, causing energy to leave the system. In our case, you observe how a calculation that begins with atmospheric pressure is later converted to energy values in joules and then to kilojoules.

Temperature and Pressure Relationships

Temperature and pressure are closely related, in that they often influence each other in gas laws. At constant temperatures, like in this exercise, Boyle's Law simplifies the relationship to only involve pressure and volume. However, in most real-world situations, a change in temperature would lead to changes in pressure and volume as well.
Understanding temperature and pressure is crucial in processes involving heating and cooling. For example, increasing the temperature of a gas in a closed container will increase its pressure, possibly leading to expansion if the container allows for it. Conversely, keeping the temperature constant necessitates maintaining an understanding of how pressure and volume adjust to adhere to gas laws.