Question

Doua cantitati identice de gaz ideal biatomic,trec din strea initiala caracterizata de parametrii p1=-10^5Pa si V1=5*10^-3 m^3 in starea finala caracterizata de parametrii p2=3*10^5 si V2= 2*10^-3 prin doua procese distincte a) prima cantitate sufera o transformare izobara urmata de una izocorasi b) a doua cantitate sufera o transformare izocora urmata de una izobara
Calculati: variatia energiei interne,caldura si lucrul mecanic??
Urgent va rog!

Answer 1

a)ΔU=υCv(T2-T1)=5/2υR(T2-T1)=5/2(p2V2-p1V1)=5/2[3*10⁵*2*10⁻³-5*10⁻³*(-10⁵)]=5/2(600+500)=5*1100/2=2750 J

Q=Q1+Q2=υCpΔT+υCvΔT

γ=7/5

Cp=γCv

Q1=υCpΔT=υγCv(T2-T1)=5*7/10*υR(T2-T1)=3,5(p2V2-p1V1)=3,5[3*10⁵*2*10⁻³-5*10⁻³*(-10⁵)]=3,5(600+500)=1100*3,5=3850 J

Q2=2750 J

Q=3850+2750=6600 J

L=p1(V2-V1)=-10⁵(2*10⁻³-5*10⁻³)=10⁵*3*10⁻³=300 J

b)

ΔU=υCv(T2-T1)=5/2υR(T2-T1)=5/2(p2V2-p1V1)=5/2[3*10⁵*2*10⁻³-5*10⁻³*(-10⁵)]=5/2(600+500)=5*1100/2=2750 J

Q=Q1-Q2=υCpΔT-υCvΔT

γ=7/5

Cp=γCv

Q1=υCpΔT=υγCv(T2-T1)=5*7/10*υR(T2-T1)=3,5(p2V2-p1V1)=3,5[3*10⁵*2*10⁻³-5*10⁻³*(-10⁵)]=3,5(600+500)=1100*3,5=3850 J

Q2=2750 J

Q=3850-2750=1100 J

L=p1(V2-V1)=-10⁵(2*10⁻³-5*10⁻³)=10⁵*3*10⁻³=300 J