Solved Problems on Thermodynamics:-Problem 1:-A container holds a mixture of three nonreacting gases: n 1 moles of the first gas with molar specific heat at constant volume C 1, and so on.Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantitites of the three separate gases. 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.

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?

  1. Thermodynamics – problems and solutions. The first law of thermodynamics. 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: Pressure (P) = 20 N/m 2. Initial volume (V 1).
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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 : Spotify deezer music downloader download.

ΔU = Q-W

ΔU = -2000-(-3000)

ΔU = -2000+3000

ΔU = 1000 Joule

Internal energy increases by 4500 Joule.
Thermodynamics
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
– If the work is done on the system, then the internal energy of the system increases

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 :

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

Thermodynamics

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

Solving Thermodynamic Problems

Wanted: Work was done by Carnot engine (W)

Solution :

The efficiency of the Carnot engine :

How To Solve Thermodynamic Problems

Work was done by Carnot engine :

Thermodynamics Problems With Solutions Pdf

W = e Q1

2500 Solved Problems In Thermodynamics Pdf Answer

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