Jawaban Tugas III

7/30/2019 Jawaban Tugas III

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TUGAS III

1. An ideal gas initially at 600 K and 10 bar undergoes a four-step mechanically reversible

cycle in a closed system. In step 12, pressure decreases isothermally to 3 bar; in step 23,

pressure decreases at constant volume to 2 bar; in step 34, volume decreases at constant

pressure; and in step 41, the gas returns adiabatically to its initial state.

a. Sketch the cycle on a PVdiagram.

b. Determine (where unknown) both Tand Pfor states 1, 2, 3, and 4.

c. Calculate Q, W, AU, and AH for each step of the cycle.

Data: Cp = (7/2)R and Cv = (5/2)R.

Penyelesaian:

a. Diagram PV:

b. 4,1R5,2R5,3

CC

V

P

R = 8,1345 J mol-1

K-1

Hubungan antara titik2 dan 3:

3

3

2

2

T

P

T

P

V

P

(bar)

10

3

2

1

2

3

4

600 K

600 K


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K4006003

2T

P

PT

2

2

3

3

Hubungan antara titik 1 dan 4:

144

1

11PTPT

K8,3782

10K600

P

PTT

4,14,111

4

1

14

c. Langkah 1-2 (proses isotermal)

U = H = 0

molJ3,587610

3lnK600

Kmol

J1345,8

P

PlnRTQ

1

2

0WQU W =Q =5876,3 J/mol

Langkah 2-3 (proses isokoris)

molJ4,2440K600400KmolJ1345,85,1TTCdTCU 1123VT

T

V

3

2

molJ3,4067K600400KmolJ1345,85,2TTCdTCH 1123PT

T

P

3

2

W = 0

QWQU

Q =2440,4 J/mol

Langkah 3-4 (proses isobaris)

molJ7,258K4008,378KmolJ1345,85,1TTCdTCU 1134VT

T

V

4

3

molJ1,431K4008,378KmolJ1345,85,2TTCdTCH 1134PT

T

P

4

3

Q = H =

431,1 J/mol

WQU

molJ6,172molJ1,4317,258QUW

Langkah 4-1 (proses adiabatis)

Q = 0


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molJ2699K8,378600KmolJ1345,85,1TTCdTCU 1141VT

T

V

1

4


T

P

1

4

WQU = W

W = U =258,7 J/mol


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2. An ideal gas, initially at 303.15 K (30C) and 100 kPa, undergoes the following cyclic processes in a closed

system:

(a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a

constant pressure of 500 kPa to 303.15 K (30C), and finally expanded isothermally to its original state.

(b)

The cycle traverses exactly the same changes of state, but each step is irreversible with an efficiency of80% compared with the corresponding mechanically reversible process.

Calculate Q, W, U, and H for each step of the process and for the cycle. Take CP= (7/2)R and CV= (5/2)R.

Penyelesaian:

R = 8,1345 J mol-1

K-1

4,1R5,2

R5,3

C

C

V

P

a.

Diagram PV:

Langkah 1-2 (proses adiabatis)

Hubungan antara titik 1 dan 2:

122

1

11PTPT

K496500

100K15,313

P

PTT

4,14,111

2

112

Q = 0

V

P

(kPa)

500

100 1

23

313,15 K

313,15 K


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T

V

2

1

molJ5,3718K15,313496KmolJ1345,85,2TTCdTCH11

12P

T

T

P

2

1

WQU = W

W = U = 2231,1 J/mol

Langkah 2-3 (proses isobaris)


T

V

3

2


T

P

3

2

Q = H =3715,5 J/mol

WQU

molJ4,1487molJ5,37181,2231QUW

Langkah 3-1 (proses isotermal)

U = H = 0

molJ8,4099

500

100lnK15,313

Kmol

J1345,8

P

PlnRTQ

3

1

0WQU

W =Q =4099,8 J/mol

Keseluruhan siklus

U = 2231,12231,1 + 0 = 0 J/mol

H = 3718,53718,5 + 0 = 0 J/mol

Q = 03718,5 + 4099,8 = 381,3 J/mol

W = 2231,1 + 1487,4

4099,8 =

381,3 J/mol


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Jawaban Tugas III

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