FIGURE 18-3
-An ideal gas undergoes the process a→b→c→a shown in Figure 18-3. P_a = P_c = 240 kPa, V_b = V_c = 40 L, V_a = 15 L, and P_b = 400 kPa. How much heat is gained by the system in this process?

Question

FIGURE 18-3 FIGURE 18-3 -An ideal gas undergoes the process a→b→c→a shown in Figure 18-3. Pa = Pc = 240 kPa, Vb = Vc = 40 L, Va = 15 L, and Pb = 400 kPa. How much heat is gained by the system in this process? A) 1000 J B) 1500 J C) 2000 J D) 2500 J E) 3000 J

FIGURE 18-3 -An ideal gas undergoes the process a→b→c→a shown in Figure 18-3. P<sub>a</sub> = P<sub>c</sub> = 240 kPa, V<sub>b</sub> = V<sub>c</sub> = 40 L, V<sub>a</sub> = 15 L, and P<sub>b</sub> = 400 kPa. How much heat is gained by the system in this process? A) 1000 J B) 1500 J C) 2000 J D) 2500 J E) 3000 J

-An ideal gas undergoes the process a→b→c→a shown in Figure 18-3. P_a = P_c = 240 kPa, V_b = V_c = 40 L, V_a = 15 L, and P_b = 400 kPa. How much heat is gained by the system in this process?

Quizplus · Accepted Answer

The process is a closed loop which means the net work done is zero. Therefore, the heat gained by the system in this process is equal to the work done on the system. Using the first law of thermodynamics, the work done on the system is given by 
W = ΔU = nCvΔT
where n is the number of moles, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature. Since the process is cyclic, the change in internal energy ΔU is also zero. Thus, the work done on the system in this process is also zero. Therefore, the heat gained by the system is also zero.

However, there seems to be an error in the problem statement as it is not possible for an ideal gas to undergo a cyclic process where the pressure and volume are constant at the same time. Therefore, the answer to this problem may not be unique or meaningful.

FIGURE 18-3 -An Ideal Gas Undergoes the Process A→b→c→a Shown in Figure