# 2500 Solved Problems In Thermodynamics Pdf

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 :

Process 1 :

Writing a good abstract for research paper pdf statistics homework 1 transportation and assignment problem docx easy topics for a process essay comparison contrast essay samples middle school what is a bibliography in research paper essay about trust and love essay diagram with triangles statement of purpose of a business plan outline examples.

• Visit the post for more. PDF 2500 Solved Problems in Fluid Mechanics and Hydraulics (Schaum’s Solved Problems) By Jack B. Evett, Cheng Liu Free Download.
• First Law of Thermodynamics 26. State the First Law of Thermodynamics. Energy can neither be created nor destroyed, only altered in form. The following schematic of a simple Rankine cycle consists of steam leaving a boiler at T=550 F and P=400 psia and passes through a turboexpander where it does work and exhausts with an enthalpy of 932.
• 2000 solved problems in thermodynamics pdf. Nursing research paper waste management business plan example how to write a creative brief for packaging business plan pro premier edition free download 2017 the best essayists homework kids doing writing coloring examples of conclusion in a research paper example.

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

2.

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

Known :

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

3.

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

Known :

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 :

Work (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

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%

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 :

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 :

Work (W) = 6000 Joule

High temperature (TH) = 800 Kelvin

Low temperature (TL) = 300 Kelvin

Wanted: heat discharged by the engine

Solution :

Carnot (ideal) efficiency :

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

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 :

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 :

High temperature (TH) = 600 Kelvin

Low temperature (TL) = 250 Kelvin

Heat input (Q1) = 800 Joule

Wanted: Work (W)

Solution :

The efficiency of Carnot engine :

Work was done by the engine :

W = e Q1

W = (7/12)(800 Joule)

W = 466.7 Joule

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 :

Work was done by Carnot engine :

W = e Q1

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

1. 3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system?

Known :

Heat (Q) = +3000 Joule

Work (W) = +2500 Joule

Wanted: the change in internal energy of the system

Solution :

The equation of the first law of thermodynamics

ΔU = Q-W

The sign conventions :

Q is positive if the heat added to the system

W is positive if work is done by the system

Q is negative if heat leaves the system

W is negative if work is done on the system

The change in internal energy of the system :

ΔU = 3000-2500

ΔU = 500 Joule

Internal energy increases by 500 Joule.

2. 2000 J of heat is added to a system and 2500 J of work is done on the system. What is the change in internal energy of the system?

Known :

Heat (Q) = +2000 Joule

Work (W) = -2500 Joule

Wanted: The change in internal energy of the system

Solution :

ΔU = Q-W

ΔU = 2000-(-2500)

ΔU = 2000+2500

ΔU = 4500 Joule

Internal energy increases by 4500 Joule.

3. 2000 J of heat leaves the system and 2500 J of work is done on the system. What is the change in internal energy of the system?

Known :

Heat (Q) = -2000 Joule

Work (W) = -3000 Joule

Wanted: The change in internal energy of the system

Solution :

ΔU = Q-W

## Thermodynamics Practice Problems Pdf

ΔU = -2000-(-3000)

ΔU = -2000+3000

ΔU = 1000 Joule

Internal energy increases by 4500 Joule.

Conclusion :

If heat is added to the system, then the internal energy of the system increases

If heat leaves the system, then the internal energy of the system decreases

If the work is done by the system, then the internal energy of the system decreases

## How To Solve Thermodynamic Problems

If the work is done on the system, then the internal energy of the system increases