# 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 :

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Pressure (P) = 20 N/m^{2}

Initial volume (V_{1}) = 10 liter = 10 dm^{3} = 10 x 10^{-3 }m^{3}

Final volume (V_{2}) = 40 liter = 40 dm^{3} = 40 x 10^{-3 }m^{3}

Process 2 :

Process (P) = 15 N/m^{2}

Initial volume (V_{1}) = 20 liter = 20 dm^{3} = 20 x 10^{-3 }m^{3}

Final volume (V_{2}) = 60 liter = 60 dm^{3} = 60 x 10^{-3 }m^{3}

__Wanted :__ The ratio of the work done by gas

__Solution :__

The work done by gas in the process I :

W = P ΔV = P (V_{2}–V_{1}) = (20)(40-10)(10^{-3 }m^{3}) = (20)(30)(10^{-3 }m^{3}) = (600)(10^{-3 }m^{3}) = 0.6 m^{3}

The work done by gas in the process II :

W = P ΔV = P (V_{2}–V_{1}) = (15)(60-20)(10^{-3 }m^{3}) = (15)(40)(10^{-3 }m^{3}) = (600)(10^{-3 }m^{3}) = 0.6 m^{3}

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

0.6 m^{3 }: 0.6 m^{3}

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 10^{5} N/m^{2} = 2 x 10^{5} Pascal

Initial volume (V_{1}) = 5 cm^{3} = 5 x 10^{-6} m^{3}

Final volume (V_{2}) = 15 cm^{3 }= 15 x 10^{-6} m^{3}

__Wanted :__ Work done by gas in process AB

__Solution :__

W = ∆P ∆V

W = P (V_{2} – V_{1})

W = (2 x 10^{5})(15 x 10^{-6 }– 5 x 10^{-6})

W = (2 x 10^{5})(10 x 10^{-6}) = (2 x 10^{5})(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 (P_{1}) = 4 Pa = 4 N/m^{2}

Final pressure (P_{2}) = 6 Pa = 6 N/m^{2}

Initial volume (V_{1}) = 2 m^{3}

Final volume (V_{2}) = 4 m^{3}

__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 10^{5} – 2 x 10^{5})

W = ½ (10)(4 x 10^{5})

W = (5)(4 x 10^{5}

W = 20 x 10^{5 }

W = 2 x 10^{6 }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 Q_{1}

6000 = (0.625) Q_{1}

Q_{1} = 6000 / 0.625

Q_{1} = 9600

Heat discharged by Carnot engine :

Q_{2} = Q_{1 }– W

Q_{2} = 9600 – 6000

Q_{2 }= 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^{o}C + 273 = 1000 K

__Wanted :__ Low temperature

__Solution :__

T_{L} = 600 Kelvin – 273 = 327^{o}C

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 (Q_{1}) = 800 Joule

__Wanted:__ Work (W)

__Solution :__

The efficiency of Carnot engine :

Work was done by the engine :

W = e Q_{1}

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 (Q_{1}) = 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 Q_{1}

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