A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
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These given conditions satisfy the second law of thermodynamics.
As the process is isobaric
So there will be a straight line of P= 200kPa in P-v and P-T planes
P1 = P2 = 100kPa
For perfect ideal gas, v-T plane:
= 287 × 500/200000 = 0.717 m³/kg
= 287 × 600/200000 = 0.861m³/kg
As it is the calorically perfect gas
de = dT
Integration on both sides
e2 - e1 = (T2 - T1)
= ( 716.5J/kg/K) (600-500)
= 71650 J/kg
also,
Tds = de + Pdv
Tds = dT +Pdv
For ideal gas
V = RT/P
dv = Rdt/P - RTdp/P²
Tds = dT + Rdt - RTdp/P
ds = ( + R)dT/T - RdP/P
ds = ()dT/T - RdP/P
ds = dT/T - RdP/P
Integration on both sides
s2 - s1 = ln (T2/T1) - R ln (P2/P1)
Since P is constant
s₂ - s₁ = ln (T2/T1)
= 1003.5 ln (600/500)
= 1003.5 × 0.182
= 182.95 J/kg/K
w = Pdv
= P(v₂ - v₁)
= 2,00,000 ( 0.861 - 0.717)
= 28,800 J/kg
de = δq -δw
δq = de + δw
q₁₂ = (e₂ - e₁) + w₁₂
= 71,650 + 28,800 = 1,00,450 J/kg
Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.
Let’s check for second law satisfaction.
s₂ - s₁ ≥ q₁₂ / Tₓ
182.95 ≥ 1,00,450 / 600 K
182.95 J/kg/K ≥ 167.41 J/kg/K
Hence this condition satisfies the second law of thermodynamics
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