Răspuns:
1-4-5 , L = L1-4 +L4-5= 0+L4-5 = niu RT1ln V5/V1 = p1V1lnV5/V1=
0,2*106*2*10-3*0,4= 0,4*0,4*1000= 160 J
L si Q sunt marimi de proces , dar U este marime de stare U1-5 = niu Cv(T5-T1)
Q= Q14+ Q4-5 = niu Cv(T4-T1)+niu Cp(T5-T4)=- niu 3/2R *200+ niu 5/2 R*100= niu R*50(-6 +5)= -p1V1/T1*50= - 0,2*2*10-3 *50/400
p1/T1=p4/T4 , p4=p5 , T4= p5T1/p1 = 0,1/0,2 *400= 200K
pt gaz monoatomic Cv = 3/2 R , Cp =5/2 R