When a gas follows path 123 on the PV diagram in the figure below, 437.0 J of he

Question

When a gas follows path 123 on the PV diagram in the figure below, 437.0 J of heat flows into the system and 177.0 J of work is done by the system.

file://localhost/Users/luckismalloy/Desktop/serw1246.gif

1)What is the change in the internal energy of the system?

2)How much heat flows into the system if the process follows path 143? The work done by the gas along this path is 61.1J.

3)What net work would be doneby the systemif the system followed path 12341?

4)What net work would be doneby the systemif the system followed path 14321?

5)What is the change in internal energy of the system in the processes described in parts c and d?

Explanation / Answer

Change of internal energy equals work done on the system plus heat transferred into the system ?U_123 = W_123 + Q_123 = -179J + 423J = 244J (Note :Work done by the system is counted negative as well as heat transferred out of the system) 2) You can describe the internal energy as function of state (p,T and V). Thus every point in you diagram represents a certain internal energy. Therefore difference of internal energy between two points does not depend on the path you follow. Hence: ?U_143 = ?U_123 = 244J Again use first law of thermodynamics: ?U_143 = W_143 + Q_143 and solve for Q_143: Q_143 = ?U_143 - W_143 = 244J - (-63.6J) = 307.6J 3) The work is given by W = - ? p dV For an isobaric step (14 and 23 and vice versa) p = constant => W = p·?V For an isochoric step (12 and 34 and vice versa) V = constant dV = 0 => W = 0 The work for the steps was given in 1) and 2). The work done following a path is the sum of the work of the single steps: W_123 = W_12 + W_23 = W_23 = -179J W_143 = W_14 + W_43 = W_14 = -63.6J If you change the direction you follow the path just change the sign: W_32 = - W_23 = 179J W_41 = - W_14 = 63.6J The reason is: Expanding from for example V_1 to V_4 the system does some work to the surrounding. You need to the same amount of work on the system to compress from V_4 to V_1. W_12341 = W_12 + W_23 + W_34 + W_41 = 0 -179J + 0 + 63.6J = -115.4 115.4 J of work are done by the system. 4.) Same as procedure as before: W_14321 = W_14 + W_43 + W_32 + W_21 = -63.6 + 0 +179J + 0 = 115.4J 115.4J of work are done ON the system following this path 5.) c and d? . I guess you think of 3) and 4). As explained in 2) the internal energy of the gas at a point in the diagram is a constant. Therefore following a cycle path, which ends up in the beginning, does not change the internal energy of the system: ?U_12341 = ?U_14321 = 0

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.