#2
W= -∫V1V2pdV | ΔU=Q+W | ΔU=($\frac{\text{σU}}{\text{σT}}$)dT+($\frac{\text{σU}}{\text{σV}}$)dV+($\frac{\text{σU}}{\text{σξ}}$)dξ | ΔH=($\frac{\text{σH}}{\text{σT}}$)dT+($\frac{\text{σH}}{\text{σp}}$)dp+($\frac{\text{σU}}{\text{σξ}}$)dξ |$\frac{\text{σU}}{\text{σV}} =$v.vals $\frac{na}{V}$ | $\frac{\text{σH}}{\text{σp}} =$szt.kul. nb | ΔH= ΔU+pdV+Vdp |
#3
ΔHr=prod-substr | ΔU=ΔH-RTΣVigaz. | ΔU, ΔHr>0 endo. Ciepło z otocz. | ΔHtw(stan wolny)=0 | ΔHr(t2)= ΔHr(t1)+∫Vi*CpdT+Vi*Hp.f+∫Vi*CpdT
#4
dS=($\frac{\text{σs}}{\text{σT}}$)dT+($\frac{\text{σS}}{\text{σV}}$)dV+($\frac{\text{σs}}{\text{σξ}}$)dξ |($\frac{\text{σs}}{\text{σT}}$)dT=$\frac{\text{Cv}}{T}$dT |($\frac{\text{σS}}{\text{σV}}$)=($\frac{\text{σp}}{\text{σT}}$)dV=nRln$\frac{V2}{V1}$(clap.)=nRln$\frac{V2 - \text{nb}}{V1 - \text{nb}}$(szt. I v.vals) | dS=($\frac{\text{σs}}{\text{σT}}$)dT+($\frac{\text{σS}}{\text{σp}}$)dV+($\frac{\text{σs}}{\text{σξ}}$)dξ |($\frac{\text{σs}}{\text{σT}}$)dT=$\frac{\text{Cp}}{T}$dT | ($\frac{\text{σS}}{\text{σp}}$)=-($\frac{\text{σV}}{\text{σT}}$)dp=-nRln$\frac{p2}{p1}$(clap.+szt.) |diS – prod.|dS – zmiana | Q/T – transp | diS=dS-Q/T | Tk=m1T1+m2T2/m1+m2 |ΔSp.f=$\frac{\text{ΣVi}*\text{Hp}.f}{\text{Tp}.f}$ |F=U-TS | dF=dU-TdS-SdT | ΔF=ΔU-Δ(TS) | G=H-TS | dG=dH-TdS-SdT | ΔG=ΔH-Δ(TS)|
#5
A=A°-RTln(p1/po)V1*(p2/po)V1 | A°=RTlnKp | A°=-ΔHr+TΔSr= -ΔGr |$\frac{\text{σlnKp}}{\text{σT}}$=$\frac{\text{ΔHr}}{RT}$=>ln$\frac{Kp2}{Kp1}$=$\frac{\text{ΔHr}}{R}$(1/T1-1/T2 | Równowaga i Tmax = A=0 |∫xndx=$\frac{1}{n + 1}x^{n + 1}$ | xn=nxn − 1 |
2.a)normal b)tabelka, wyliczenie x | 3.Cp/tdT-nR*lnp2/p1 | 4.T2
1.298273200 |2.b) A=A°-RTln(pc*x1/po)V1*(pc*x1/po)V2*(pc*x3/po)V3
2.pzew=p2 | 3.ds=cp/tdT-nrlnp2/p1, s=s298+ncp/tdT | 4.V=1 p=nrt/v, c)Kp=(xno2)2*(xn2o4)-1 , 1-xno2=xn2o4 , -xno2^2-Kpxno2+kp=0
#2
W= -∫V1V2pdV | ΔU=Q+W | ΔU=($\frac{\text{σU}}{\text{σT}}$)dT+($\frac{\text{σU}}{\text{σV}}$)dV+($\frac{\text{σU}}{\text{σξ}}$)dξ | ΔH=($\frac{\text{σH}}{\text{σT}}$)dT+($\frac{\text{σH}}{\text{σp}}$)dp+($\frac{\text{σU}}{\text{σξ}}$)dξ |$\frac{\text{σU}}{\text{σV}} =$v.vals $\frac{na}{V}$ | $\frac{\text{σH}}{\text{σp}} =$szt.kul. nb | ΔH= ΔU+pdV+Vdp |
#3
ΔHr=prod-substr | ΔU=ΔH-RTΣVigaz. | ΔU, ΔHr>0 endo. Ciepło z otocz. | ΔHtw(stan wolny)=0 | ΔHr(t2)= ΔHr(t1)+∫Vi*CpdT+Vi*Hp.f+∫Vi*CpdT
#4
dS=($\frac{\text{σs}}{\text{σT}}$)dT+($\frac{\text{σS}}{\text{σV}}$)dV+($\frac{\text{σs}}{\text{σξ}}$)dξ |($\frac{\text{σs}}{\text{σT}}$)dT=$\frac{\text{Cv}}{T}$dT |($\frac{\text{σS}}{\text{σV}}$)=($\frac{\text{σp}}{\text{σT}}$)dV=nRln$\frac{V2}{V1}$(clap.)=nRln$\frac{V2 - \text{nb}}{V1 - \text{nb}}$(szt. I v.vals) | dS=($\frac{\text{σs}}{\text{σT}}$)dT+($\frac{\text{σS}}{\text{σp}}$)dV+($\frac{\text{σs}}{\text{σξ}}$)dξ |($\frac{\text{σs}}{\text{σT}}$)dT=$\frac{\text{Cp}}{T}$dT | ($\frac{\text{σS}}{\text{σp}}$)=-($\frac{\text{σV}}{\text{σT}}$)dp=-nRln$\frac{p2}{p1}$(clap.+szt.) |diS – prod.|dS – zmiana | Q/T – transp | diS=dS-Q/T | Tk=m1T1+m2T2/m1+m2 |ΔSp.f=$\frac{\text{ΣVi}*\text{Hp}.f}{\text{Tp}.f}$ |F=U-TS | dF=dU-TdS-SdT | ΔF=ΔU-Δ(TS) | G=H-TS | dG=dH-TdS-SdT | ΔG=ΔH-Δ(TS)|
#5
A=A°-RTln(p1/po)V1*(p2/po)V1 | A°=RTlnKp | A°=-ΔHr+TΔSr= -ΔGr |$\frac{\text{σlnKp}}{\text{σT}}$=$\frac{\text{ΔHr}}{RT}$=>ln$\frac{Kp2}{Kp1}$=$\frac{\text{ΔHr}}{R}$(1/T1-1/T2 | Równowaga i Tmax = A=0 |∫xndx=$\frac{1}{n + 1}x^{n + 1}$ | xn=nxn − 1 |
2.a)normal b)tabelka, wyliczenie x | 3.Cp/tdT-nR*lnp2/p1 | 4.T2
1.298273200 |2.b) A=A°-RTln(pc*x1/po)V1*(pc*x1/po)V2*(pc*x3/po)V3
2.pzew=p2 | 3.ds=cp/tdT-nrlnp2/p1, s=s298+ncp/tdT | 4.V=1 p=nrt/v, c)Kp=(xno2)2*(xn2o4)-1 , 1-xno2=xn2o4 , -xno2^2-Kpxno2+kp=0