Consider the process shown on the PV diagram to the right. In this process, 1.20

Question

Consider the process shown on the PV diagram to the right. In this process, 1.20 moles of a monatomic ideal gas perform a complete cycle from the initial state A to state B, then to state C, then to state D, and then back to state A. (Note that the pressure axis is multiplied by 10^5 Pa and the volume axis by 10^-3 m^3.) (A). Calculate the temperature of the gas at A, B, C, and D. Show the detail for one point! (B). Calculate the change in the internal energy of the gas during the step from A to B: B to C: C to D: D to A. Show the detail from A to B. What is the change in internal energy over one complete cycle (from A back to A)?

Explanation / Answer

now we find the temperature at A,B,C and D

Ta=5*10^5*6*10^-3/8.3= 361.45 K

Tb=5*10^5*2*10^-3/8.3=120.5 k

Tc=10*2*10^5*10^-3/8.3=240.96 K

Td=10*6*10^5*10^-3/8.3=602.41 k

now we find the internal energy in each case

internal energy UAB=1.2*3*8.3*[120.5-361.45]/2=-3599.8 J

internal energy UBC=1.2*3*8.3*[240.96-361.45]/2=-1800.1 J

internal energy UCD=1.2*8.3*3(602.41-240.96)/2=5400.1 J

nternal energy UDA=1.2*8.3*3*(361.45-602.41)/2=-3599.9 J

the total internal energy U=-3599.9+5400.1-1800.1-3599.8=-3599.7 J

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