1. Thermodynamics Solution Manual Pdf
2. Thermodynamics Problems And Solutions
3. Thermodynamics Questions And Solutions 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?

Thermodynamics key facts (7/9). Ideal gas law. 1. Form: 𝑃𝑃𝑉𝑉= 𝑁𝑁𝑘𝑘. 𝑇𝑇. 𝑃𝑃=Pressure, 𝑉𝑉=Volume, 𝑁𝑁=number of molecules, 𝑘𝑘. 𝐵𝐵 =Boltzmann’s constant, 𝑇𝑇=temperature in K. 2. Form: 𝑃𝑃𝑉𝑉= 𝑛𝑛𝑅𝑅𝑇𝑇. 𝑛𝑛=number. Problem: Given that the free energy of formation of liquid water is -237 kJ / mol, calculate the potential for the formation of hydrogen and oxygen from water. To solve this problem we must first calculate ΔG for the reaction, which is -2 ( -237 kJ / mol) = 474 kJ / mol. Knowing that ΔG = -nFE o and n = 4, we calculate the potential is -1.23 V. Thermodynamics problems and solutions When solving a Physics problem in general and one of Thermodynamics in particular, it is important to follow a certain order. Get used to being organized when you solve problems, and you will see how it gives good results. Processes (Ideal Gas) A steady flow compressor handles 113.3 m 3 /min of nitrogen (M = 28; k = 1.399) measured at intake where P1= 97 KPa and T1= 27 C. Discharge is at 311 KPa. The changes in KE and PE are negligible. For each of the following.

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

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 (Q_H) = 2000 Joule

Heat output (Q_L) = 1200 Joule

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

Wanted : efficiency (e)

Solution :

e = W / Q_H

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 (T_H) = 960 K

Low temperature (T_L) = 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 (T_H) = 800 Kelvin

Low temperature (T_L) = 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 (T_H) = 727^oC + 273 = 1000 K

Wanted : Low temperature

Solution :

T_L = 600 Kelvin – 273 = 327^oC

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

Known :

High temperature (T_H) = 600 Kelvin

Low temperature (T_L) = 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 (T_L) = 400 K

High temperature (T_H) = 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. Two kg of air at 500kPa, 80°C expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100kPa and 5°C.

For this process, determine

1)the maximum work

2)the change in a availability and

3)the irreversibility.

Take, C_v = 0.718 KJ/kg K, R = 0.287 KJ/kg K.

2.A reversible heat engine receives 3000 KJ of heat from a constant temperature source at 650 K . If the surroundings is at 295 K,

determine

i)the availability of heat energy

ii)Unavailable heat.

A = Q₁ –T₀ (ΔS)

=3000 –295 (3.17)

=2064.85 KJ.

Unavailable heat (U.A) = T₀ (ΔS)

=295 (3.17)

=935.15 KJ.

Result:

1)The availability of heat energy (A) = 2064.85 KJ.

2)Unavailable heat (U.A) = 935.15 KJ.

3. Air in a closed vessel of fixed volume 0.15 m³ exerts pressure of 12 bar at 250 °C. If the vessel is cooled so that the pressure falls to 3.5 bar , determine the final pressure, heat transfer and change of entropy.

Given Data:

V₁ = 0.15 m³

p₁ = 12 bar = 1200 KN/m² p₂ = 3.5 bar = 350 KN/m²

T₁ = 250°C = 273+250 = 523 K

To find:

1)The final pressure,

2)Heat transfer

3)Change of entropy.

Solution:

Result:

Thermodynamics Solution Manual Pdf

1)The final pressure, T₂ = 152.54 K.

2)Heat transfer, Q = - 446.78 KJ.

3) Change of entropy,-1.06KJ/K. ΔS =

4. A domestic food freezer maintains a temperature of - 15°C. The ambient air is at 30°C. If the heat leaks into the freezer at a continuous rate of 1.75 KJ/s, what is the least power necessary to pump the heat out continuously?

Given Data:

T_L = - 15°C = 273 –15 = 258 K

T_H = 30°C = 273 + 30 = 303 K

Q_S = 1.75 KW

To find:

Least power, (W)

Solution:

Result:

Least power necessary to pump heat, W = 0.305 KW.

5. A refrigerator working on reversed Carnot cycle requires 0.5 KW per KW of cooling to maintain a temperature of -15°C. Determine the following:

a)COP of the refrigerator

b)Temperature at which heat is rejected and

Amount of heat rejected to the surroundings per KW of cooling.

Given Data:

W = 0.5 KW

Q₂ = 1 KW

T₂ = -15°C = 273 –15 = 258 K.

To find:

1)COP of the refrigerator (COP)

2)Temperature at which heat is rejected (T₁)

3)Amount of heat rejected to the surroundings per KW of cooling (Q₁)

Solution:

Thermodynamics Problems And Solutions

Result:

1)COP of the refrigerator (COP) = 2

2)Temperature at which heat is rejected (T₁) = 387 K

Amount of heat rejected to the surroundings per KW of cooling (Q₁) = 1.5 KW.

Thermodynamics Questions And Solutions Pdf