Futaba21
Dik. n = 2,4 mol T1 = 47°C = 320 K T2 = 3. T1 = 3. 320 = 960 K ΔT = T2 - T1 = 640 K R = 8,3 J mol^−1 K^−1 Dit. W............?
penyelesaian : gas diatomik : ΔU = 5/2. n. R. ΔT ΔU = 5/2. 2.4 . 8.3 . 640 ΔU = 3.1872x10^4 Joule melalui proses adiabatik ΔU = W maka, W = ΔU W = 3.1872x10^4 Joule
n = 2,4 mol
T1 = 47°C = 320 K
T2 = 3. T1 = 3. 320 = 960 K
ΔT = T2 - T1 = 640 K
R = 8,3 J mol^−1 K^−1
Dit. W............?
penyelesaian :
gas diatomik :
ΔU = 5/2. n. R. ΔT
ΔU = 5/2. 2.4 . 8.3 . 640
ΔU = 3.1872x10^4 Joule
melalui proses adiabatik ΔU = W maka,
W = ΔU
W = 3.1872x10^4 Joule
koreksi kembali bila keliru ^^
• Usaha oleh gas
Proses adiabatik
gas diatomik
n = 2,4 mol
T1 = 47 °C = 320 K
T2 = 3 T1 = 960 K
W = __?
CARA 1
Persamaan keadaan
P1 V1 = n R T1
P2 V2 = n R T2 = 3 n R T1
Usaha
W = 1/(γ-1) {P1 V1 - P2 V2}
W = 1/(1,4-1) {n R T1 - 3 n R T1}
W = 1/(0,4) {-2 n R T1}
W = 10/4 {-2 (2,4) (8,3) (320)}
W = - 31872 J ← jwb
CARA 2
Perubahan Energi dalam
∆U = 5/2 n R (T2 - T1)
Usaha (adiabatik, Q = 0)
∆U = Q - W
∆U = 0 - W
W = - ∆U
W = - 5/2 n R (T2 - T1)
W = - 5/2 (2,4) (8,3) (960-320)
W = - 31872 J ← jwb