• 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.
  • The following two statements of the second law of thermodynamics are based on the definitions of the heat engines and heat pumps. Kelvin-Planck statement of the second law It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.

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?

Download Thermodynamics Problems With Solutions book pdf free download link or read online here in PDF. Read online Thermodynamics Problems With Solutions book pdf free download link book now. All books are in clear copy here, and all files are secure so don't worry about it.

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

Problems

W is positive if work is done by the system

Thermodynamics Chemistry Problems And Solutions Pdf

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

Physics Problems And Solutions Pdf

Work (W) = -2500 Joule

Wanted: The change in internal energy of the system

Solution :

ΔU = Q-W

ΔU = 2000-(-2500)

ΔU = 2000+2500

Patch fr sims 3 complete edition. 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

Thermodynamics Problems And Solutions Pdf

Work (W) = -3000 Joule

Wanted: The change in internal energy of the system

Solution :

ΔU = Q-W

ΔU = -2000-(-3000)

ΔU = -2000+3000

ΔU = 1000 Joule

Internal energy increases by 4500 Joule.

Conclusion :

Thermodynamics Problems And Solutions Pdf Free

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

Physics Thermodynamics Problems And Solutions Pdf

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