Mohon bantuannya (14)……

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 = 3.1872×10^4 Joule 
melalui proses adiabatik ΔU = W maka, 
W = ΔU 
W = 3.1872×10^4 Joule 

koreksi kembali bila keliru ^^

TERMODINAMIKA
• 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