Solve
Asked by g_archanasharma
28th April 2018,
8:58 AM
Answered by Expert
Answer:
for mono-atomic gas :- Cv = (3/2)R, Cp = (5/2)R
for di-atomic gas :- Cv = (5/2)R, Cp = (7/2)R
for a mixture of gas each property should be multipled by weight factor and then added together. Mole fraction is the weight factor.
Specific heat of mixture at constant pressure Cp :- (2/5)×(7/2)R + (3/5)×(5/2)R
Specific heat of mixture at constant volume Cv :- (2/5)×(5/2)R + (3/5)×(3/2)R
Then Cp/Cv = 29/19
Answered by Expert
28th April 2018,
11:50 AM
Rate this answer
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
You have rated this answer /10
Your answer has been posted successfully!
RELATED STUDY RESOURCES :
Browse free questions and answers by Chapters
- 1 Thermodynamics
- 2 Gravitation
- 3 Electromagnetic Waves
- 4 Communication Systems
- 5 Laws of Motion
- 6 Current Electricity
- 7 Work, Energy and Power
- 8 Kinematics
- 9 Physics and Measurement
- 10 Rotational Motion
- 11 Properties of Solids and Liquids
- 12 Kinetic Theory of Gases
- 13 Oscillations and Waves
- 14 Electrostatics
- 15 Magnetic Effects of Current and Magnetism
- 16 Electromagnetic Induction and Alternating Currents
- 17 Optics
- 18 Dual Nature of Matter and Radiation
- 19 Atoms and Nuclei
- 20 Electronic Devices
Trending Tags
Latest Questions
JEE Main Mathematics
Asked by amaladelsingh
28th October 2020,
10:24 PM
CBSE X Biology
Asked by reenacharan2008
28th October 2020,
9:11 PM
