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Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ