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An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, pressure increases isothermally to 5 bar; in step 23, pressure increases at constant volume; and in step 31, the gas returns adiabatically to its initial state. Take CP=(7/2)R and CV=(5/2)R. (a) Sketch the cycle on a PV diagram. (b) Determine (where unknown) V, T, and P for states 1, 2, and 3. (c) Calculate Q, W, ΔU, and ΔH for each step of the cycle.

Respuesta :

shadrachadamu

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

Ver imagen shadrachadamu
Ver imagen shadrachadamu
Ver imagen shadrachadamu
Ver imagen shadrachadamu
lhmarianateixeira

In this exercise we have to use the knowledge of closed system to calculate some characteristics of this system, like this:

A)-4014.26

B) 569.5 K

C) -5601.55

Organizing the information given in the exercise we have:

  • An ideal gas initially at 300 K
  • 1 bar undergoes a three-step
  • increases isothermally to 5 bar
  • Take CP=(7/2)R and CV=(5/2)R

A)So doing the calculation of the PV cycle we have:

[tex]Q12-W12 = -4014.26\\AU2s = Q23= 5601.55\\AUs = Ws = -5601.55\\=-4014.26+5601.55-5601.5=-4014.26[/tex]

B) Calculation of the P, V and T for all the three states, we have:

[tex]T_2=T_3(P_2/P_3)= 569.5K\\P_3=P_1(V_1/V_3)=9.49 bar[/tex]

C) Calculation of Q, W and U for each step, we have:

  • Work will be:

[tex]Wi_2= RTi ln\\8.314 * 300 In =4014.26[/tex]

  • Amount of heat:

[tex]AU12= W12 +Q12\\Q12-W12 = -4014.26[/tex]

  • Internal energy:

[tex]\Delta U=C, (T_i-T)= 8.314 (300)\\\Delta U = W = -5601.55[/tex]

See more about termodinamics at brainly.com/question/8987993

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