Look at the P-V diagram below (Diagram 1). Calculate the work done by the gas for the paths A, B and C. Assume that in Diagram 1, P1 = 1 atm, P2 = 4 atm, V1 = 5 L, V2 = 15 L.a) WA = 1013 J, WB = 0, WC = -2533 Jb) WA = 0.01 J, WB = 0, WC = -0.025 J c) WA = 2533 J, WB = 0, WC = -1013 JCalculate the work done by the gas for the path AB in Diagram 2. Use the data: P1 = 1 atm, P2 = 4 atm, V1 = 5 L, V2 = 20 L. (Path AB is an "isothermal" which means the temperature T is constant on this path).a) 0.012 Jb) 1220 J c) 0.0278 J d) 2809 J The image presents two graphs side by side.**Left Graph:**- **Axes:** - Y-axis: Amplitude - X-axis: Time- **Description:** - The graph is a triangle wave, which starts at the origin and slopes upwards. It forms a triangular shape with an ascending line, then a horizontal line, and finally, a descending line back to the baseline. - The Time on the X-axis progresses evenly, while the Amplitude on the Y-axis increases to a peak and symmetrically decreases.**Right Graph:**- **Axes:** - Y-axis: Intensity - X-axis: Distance- **Description:** - This graph depicts an inverse square law relationship, which is typically characterized by a curve that decreases sharply. - As the Distance on the X-axis increases, the Intensity on the Y-axis decreases rapidly, illustrating the principle that intensity diminishes with the square of the distance from the source.These graphs visually represent important principles in physics: the dynamic nature of wave amplitudes over time, and the inverse square law of intensity as it relates to distance.

Dear student, According to the guidelines I am allowed to answer only one problem in a single post and you didn't mention which problem's solution you required that's why I have solve first problem. Please upload the remaining problems in another post one by one.1 Atm = 101325 Pa 1L = 10-3 m3 Write the equation for the work done. W = Pressure *change in volume = P(vf - vi) Calculate the work done in process A. WA = P1(v2 -v1) = 101325*(15-5)*10-3 WA = 1013.2 J The change in volume in the process B is zero so WB = 0 J The work done by the process A and B are correctly match with only option (a). Hence, the correct option is (a).