In this scenario, we analyze the behavior of a monoatomic gas undergoing two distinct thermodynamic processes: an isobaric process followed by an isovolumetric process. Initially, the gas is at a temperature of 293 Kelvin. The first process involves adding 2,000 joules of heat at constant pressure, which results in a temperature increase. The second process entails removing the same amount of heat at constant volume, leading to a temperature decrease. Our goal is to determine the final temperature of the gas after these processes.

To visualize these processes, we can represent them on a PV diagram. The transition from point A to point B, where heat is added at constant pressure, appears as a horizontal line (isobaric process). The transition from point B to point C, where heat is removed at constant volume, is depicted as a vertical line moving downward (isovolumetric process).

We denote the initial temperature at point A as \( T_A = 293 \, \text{K} \). The final temperature at point C, \( T_C \), can be expressed as:

\[ T_C = T_A + \Delta T_{AB} + \Delta T_{BC} \]

Here, \( \Delta T_{AB} \) represents the temperature change during the isobaric process, and \( \Delta T_{BC} \) represents the temperature change during the isovolumetric process.

For the isobaric process (A to B), we use the equation:

\[ q = n C_P \Delta T_{AB} \]

Where \( q \) is the heat added, \( n \) is the number of moles, and \( C_P \) is the molar specific heat at constant pressure. Given that \( q = 2000 \, \text{J} \) and \( n = 3 \) moles, we can calculate \( C_P \) for a monoatomic gas as:

\[ C_P = \frac{5}{2} R \]

Substituting \( R = 8.314 \, \text{J/(mol K)} \), we find:

\[ C_P = \frac{5}{2} \times 8.314 = 20.785 \, \text{J/(mol K)} \]

Now, rearranging the equation to solve for \( \Delta T_{AB} \):

\[ \Delta T_{AB} = \frac{q}{n C_P} = \frac{2000}{3 \times 20.785} \approx 32.1 \, \text{K} \]

Next, for the isovolumetric process (B to C), we use the equation:

\[ q = n C_V \Delta T_{BC} \]

Here, \( C_V \) is the molar specific heat at constant volume, calculated as:

\[ C_V = \frac{3}{2} R \]

Substituting \( R \) gives:

\[ C_V = \frac{3}{2} \times 8.314 = 12.471 \, \text{J/(mol K)} \]

Since we are removing heat, \( q \) is negative, so:

\[ \Delta T_{BC} = \frac{-2000}{3 \times 12.471} \approx -53.5 \, \text{K} \]

Now, we can calculate the final temperature:

\[ T_C = 293 + 32.1 - 53.5 \approx 271.6 \, \text{K} \]

This result indicates that the final temperature of the gas after both processes is 271.6 Kelvin. The analysis highlights how the specific heat capacities influence temperature changes during different thermodynamic processes, with the isovolumetric process resulting in a greater temperature change for the same amount of heat transfer compared to the isobaric process.