p=constanta , legea Gay-lusac : V1/T1=V2/ T2, dar V2=zV1, obtinem T2=zT1 , T1=17+273= 290K , T2-T1=T1(z-1)
Q=mc (T2-T1) =0.014 . T1(k-1) .c , caldura specifica trebuie luata dintr-un tabel pt azot
L=p(V2-V1)=p(z-1)V1.
din ecuatia termica de stare pV=m/miuRT , miu =14
U2-U!=Q-L