Question

Consider the following isobaric process for air, modelled as a Calorically Perfect Ideal Gas

Answer 1

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:

$v = (\frac{R}{P}) T$

$v_{1} = (\frac{R}{P_{1} }) T_{1}$ = 287 × 500/200000 = 0.717 m³/kg

$v_{2} = (\frac{R}{P_{2} }) T_{2}$ = 287 × 600/200000 = 0.861m³/kg

As it is the calorically perfect gas

de = $c_{v}$dT

Integration on both sides

e2 - e1 = $c_{v}$(T2 - T1)

           = ( 716.5J/kg/K) (600-500)

           = 71650 J/kg

also,

Tds = de + Pdv

Tds = $c_{v}$dT +Pdv

For ideal gas

V = RT/P        

dv = Rdt/P - RTdp/P²

Tds = $c_{v}$dT + Rdt - RTdp/P

ds = ($c_{v}$ + R)dT/T - RdP/P

ds = ($c_{v} + c_{p} -c_{v}$)dT/T - RdP/P

ds = $c_{p}$dT/T - RdP/P

Integration on both sides

s2 - s1 = $c_{p}$ln (T2/T1) - R ln (P2/P1)

Since P is constant

s₂ - s₁ = $c_{p}$ ln (T2/T1)

           = 1003.5 ln (600/500)

           = 1003.5 × 0.182

           = 182.95 J/kg/K

w = Pdv

$w_{12}$ = 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

Learn more about laws of thermodynamics here brainly.com/question/13164851

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