a) Q=ΔU+L
calculam nr de moli: ν(niu) =[tex] \frac{m}{M} [/tex]=[tex] \frac{14*10^{-3} kg}{28*10^{-3} kg/mol} [/tex]= 0,5 moli.
ΔU=niu * Cv(T2-T1).
intram in legea transformarii izobare: [tex] \frac{V1}{T1}= \frac{V2}{T2} => T2= \frac{T1*V2}{V1}= \frac{T1* 1,2V1}{V1}= 1,2*T1 [/tex]
ΔU=niu* Cv(1,2T1-T1)=0,5*[tex] \frac{5}{2} * 8,31[/tex] * 0,2*290=602,47 J
L=p(V2-V1)=p(1,2*V1-V1)=p*0,2V1=0,2*niu*R*T=0,2*0,5*8,31*290=240,99 J
Q=602,47+240,99=843,46 J
b)L=240,99 J - l-am calculat mai devreme
c) ΔT=T2-T1= 1,2*T1-T1=0,2*T1=0,2*290=58 K
