11,.A Carnot heat engine contains 8.00 moles of oxygen gas (assume Cv.m= 2.5R). Thetemperature of the heat source is 500.0 °C and the temperature of the cold reservoir is 25.0 °C.Calculate the thermodynamic efficiency, V₁, V2, V3, V4, P3, P4, w and q for each step of the cycleand the total heat and work of the cycle if P1 = 20.0 bar and P₂ = 5.00 bar.

Answered: 11. ₁. A Carnot heat engine contains…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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11,
.A Carnot heat engine contains 8.00 moles of oxygen gas (assume Cv.m= 2.5R). The
temperature of the heat source is 500.0 °C and the temperature of the cold reservoir is 25.0 °C.
Calculate the thermodynamic efficiency, V₁, V2, V3, V4, P3, P4, w and q for each step of the cycle
and the total heat and work of the cycle if P1 = 20.0 bar and P₂ = 5.00 bar.

Expert Solution

Step 1

A carnot cycle consists of 4 steps.

1) reversible isothermal expansion (P₁,V₁,T_H to P₂, V₂,T_H)

2) reversible adiabatic expansion (P₂, V₂, T_H to P₃, V₃, T_C)

3) reversible isothermal compression (P₃, V₃, T_C to P₄, V₄, T_C)

4) reversible adiabatic compression (P₄, V₄, T_C to P₁, V₁, T_H)

P₁ = 20 bar = 20x0.987 atm = 19.74 atm {1 bar = 0.987 atm)

P₂ = 5bar = 5x 0.987atm = 4.935 atm

T_H = 500^oc = 500+273 = 773 K

T_C = 25^oc = 25+273 = 298K

Cv = 2.5R = (5/2) R

Cp = (7/2) R

$γ$ = Cp/Cv = 7/5

It follows ideal gas equation

P₁ V₁ = nR T_H

V₁ = (nR T_H)/ P₁ = (8mole x 0.082 L atm mol-1 K-1 x 773 K) / 19.74 atm = 25.69 L

V₂ = (nR T_H) / P₂ = (8mole x 0.082 L atm mol-1 K-1 x 773 K) / 4.935 atm = 102.75 L

again T_H ${V_{2}}^{γ - 1}$ = Tc ${V_{3}}^{γ - 1}$ (as adiabatic process)

V₃ = ${(\frac{T_{H}}{T_{C}})}^{1 - γ}$ x V₂ = ${(\frac{773}{298})}^{(- \frac{2}{5})} \times 102.75$ = 70.18 L

So, P₃ = $\frac{n R T_{C}}{V_{3}}$ = $\frac{8 \times 0.082 \times 298}{70.18}$ = 2.785 atm

again,

$T_{C} \times {(V_{4})}^{γ - 1} = T_{H} \times {(V_{1})}^{γ - 1} V_{4} = {(\frac{T_{H}}{T_{C}})}^{1 - γ} \times V_{1} = {(\frac{773}{298})}^{(- \frac{2}{5})} \times 25.69 = 17.54 L$ (as adiabatic process)

$P_{4} = (\frac{n R T_{C}}{V_{4}}) = (\frac{8 \times 0.082 \times 298}{17.54}) = 11.14 a t m$

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