Thermodynamics – problems and solutions

The first law of thermodynamics

1. Based on graph P-V below, what is the ratio of the work done by the gas in the process I, to the work done by the gas in the process II?

Known :Thermodynamics – problems and solutions 1

Process 1 :

Pressure (P) = 20 N/m2

Initial volume (V1) = 10 liter = 10 dm3 = 10 x 10-3 m3

Final volume (V2) = 40 liter = 40 dm3 = 40 x 10-3 m3

Process 2 :

Process (P) = 15 N/m2

Initial volume (V1) = 20 liter = 20 dm3 = 20 x 10-3 m3

Final volume (V2) = 60 liter = 60 dm3 = 60 x 10-3 m3

Wanted : The ratio of the work done by gas

Solution :

The work done by gas in the process I :

W = P ΔV = P (V2–V1) = (20)(40-10)(10-3 m3) = (20)(30)(10-3 m3) = (600)(10-3 m3) = 0.6 m3

The work done by gas in the process II :

W = P ΔV = P (V2–V1) = (15)(60-20)(10-3 m3) = (15)(40)(10-3 m3) = (600)(10-3 m3) = 0.6 m3

The ratio of the work done by gas in the process I and the process II :

0.6 m3 : 0.6 m3

1 : 1

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2.

Based on the graph below, what is the work done by helium gas in the process AB?

Thermodynamics – problems and solutions 2Known :

Pressure (P) = 2 x 105 N/m2 = 2 x 105 Pascal

Initial volume (V1) = 5 cm3 = 5 x 10-6 m3

Final volume (V2) = 15 cm3 = 15 x 10-6 m3

Wanted : Work done by gas in process AB

Solution :

W = ∆P ∆V

W = P (V2 – V1)

W = (2 x 105)(15 x 10-6 – 5 x 10-6)

W = (2 x 105)(10 x 10-6) = (2 x 105)(1 x 10-5)

W = 2 Joule

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3.

Based on the graph below, what is the work done in process a-b?

Thermodynamics – problems and solutions 3Known :

Initial pressure (P1) = 4 Pa = 4 N/m2

Final pressure (P2) = 6 Pa = 6 N/m2

Initial volume (V1) = 2 m3

Final volume (V2) = 4 m3

Wanted : work done I process a-b

Solution :

Work done by gas = area under curve a-b

W = area of triangle + area of rectangle

W = ½ (6-4)(4-2) + 4(4-2)

W = ½ (2)(2) + 4(2)

W = 2 + 8

W = 10 Joule

4. Based on graph below, what is the work done in process A-B-C-A.

Solution :

Thermodynamics – problems and solutions 4Work (W) = Area of the triangle A-B-C

W = ½ (20-10)(6 x 105 – 2 x 105)

W = ½ (10)(4 x 105)

W = (5)(4 x 105)

W = 20 x 105

W = 2 x 106 Joule

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Heat engine

5. An engine absorbs 2000 Joule of heat at a high temperature and exhausted 1200 Joule of heat at a low temperature. What is the efficiency of the engine?

Known :

Heat input (QH) = 2000 Joule

Heat output (QL) = 1200 Joule

Work done by engine (W) = 2000 – 1200 = 800 Joule

Wanted : efficiency (e)

Solution :

e = W / QH

e = 800/2000

e = 0.4 x 100%

e = 40%

Carnot engine

6. An engine absorbs heat at 960 Kelvin and the engine discharges heat at 576 Kelvin. What is the efficiency of the engine.

Known :

High temperature (TH) = 960 K

Low temperature (TL) = 576 K

Wanted: efficiency (e)

Solution :

Thermodynamics – problems and solutions 5

Efficiency of Carnot engine = 0.4 x 100% = 40%

7. Based on the graph below, work done by the engine is 6000 Joule. What is the heat discharged by engine each circle?

Known :Thermodynamics – problems and solutions 6

Work (W) = 6000 Joule

High temperature (TH) = 800 Kelvin

Low temperature (TL) = 300 Kelvin

Wanted: heat discharged by the engine

Solution :

Carnot (ideal) efficiency :

Thermodynamics – problems and solutions 7

Heat absorbed by Carnot engine :

W = e Q1

6000 = (0.625) Q1

Q1 = 6000 / 0.625

Q1 = 9600

Heat discharged by Carnot engine :

Q2 = Q1 – W

Q2 = 9600 – 6000

Q2 = 3600 Joule

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8. The efficiency of a Carnot engine is 40%. If heat absorbed at 727°C then what is the low temperature.

Known :

Efficiency (e) = 40% = 40/100 = 0.4

High temperature (TH) = 727oC + 273 = 1000 K

Wanted : Low temperature

Solution :

Thermodynamics – problems and solutions 8

TL = 600 Kelvin – 273 = 327oC

9. Based on graph below, if the engine absorbs 800 J of heat, what is the work done by the engine.

Known :Thermodynamics – problems and solutions 9

High temperature (TH) = 600 Kelvin

Low temperature (TL) = 250 Kelvin

Heat input (Q1) = 800 Joule

Wanted: Work (W)

Solution :

The efficiency of Carnot engine :

Thermodynamics – problems and solutions 10

Work was done by the engine :

W = e Q1

W = (7/12)(800 Joule)

W = 466.7 Joule

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10. The high temperature of a Carnot engine is 600 K. If the engine absorbs 600 J of heat and the low temperature is 400 K, what is the work done by the engine.

Known :

Low temperature (TL) = 400 K

High temperature (TH) = 600 K

Heat input (Q1) = 600 Joule

Wanted: Work was done by Carnot engine (W)

Solution :

The efficiency of the Carnot engine :

Thermodynamics – problems and solutions 11

Work was done by Carnot engine :

W = e Q1

W = (1/3)(600) = 200 Joule