**PV Diagram of Gas Transformation****Initial and Final Conditions:**- Initially, five moles of gas are at a pressure of 2.00 atm and a volume of 0.300 L with an internal energy of 91.0 J.- In the final state, the gas reaches a pressure of 1.50 atm and a volume of 0.800 L, with the internal energy equaling 182 J.**Diagram Explanation:**The plot is a Pressure (P) vs. Volume (V) graph showing different paths of transformation:1. **Path IAF**: - Path connects points I to A in a horizontal line, then A to F in a vertical line.2. **Path IBF**: - Path connects points I to B at a diagonal, then B to F in a vertical line.3. **Path IF**: - Path directly connects points I to F diagonally.**Work Done Calculation:**- **\( W_{IAF} = 75.99 \, \text{J} \)** (Incorrect sign)- **\( W_{IBF} = 101.325 \, \text{J} \)** (Incorrect sign)- **\( W_{IF} = 88.65 \, \text{J} \)** (Incorrect sign)Each path represents a different sequence of pressure and volume changes, indicating how the work done varies with each path. Correct calculation requires considering the sign of work done, with the given responses having incorrect signs.

The work done by the system on a path in the PV- diagram is the value of the area under the PV…

Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J. P (atm) 2.00- 1.50 V (liters) 0.300 0,800 (a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. 75.99 The response you submitted has the wrong sign. J 101.325 W IBF WJAF= The response you submitted has the wrong sign. J 88.65 I The response vou submitted has the wrong sign 1

Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J. P (atm) 2.00- 1.50 V (liters) 0.300 0,800 (a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. 75.99 The response you submitted has the wrong sign. J 101.325 W IBF WJAF= The response you submitted has the wrong sign. J 88.65 I The response vou submitted has the wrong sign 1

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J. P (atm) 2.00 B A 1.50 F V (liters) 0.300 0.800 (a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. W IAF = W = IBF W = IF J (b) For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process. QIAF = QIBF QIF = J J J