T= 27°C= 27+273
P= 1atm
V1= 10dm3
V2= 20dm3
1. q= 0, for an adiabatic expansion, heat is neither gained nor lost by the system.
2. w= -P(V2-V1)
w= -1(20-10)
w= -10L.atm
1L.atm= 101.3J
-10L.atm= -10 x 101.3J= -1013J
3. ∆U = q+w
Since q= 0 (adiabatic)
∆U=w
∆U=-1013J
4. ∆H=∆U+P∆V
∆H=-1013+1(20-10)
∆H= -1013+10
∆H= -1003J
5. V1= 10dm3 T1= 300K
V2= 20dm3 T2= ?
V1/T1 = V2/T2
T2= V2T1/V1 = 20x300/10
= 600K = 327°C
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