Questions
6 moles of oxygen, 4 moles of Ar. and 2 moles of water vapour form an ideal gas mixture. The mixture is made to expand adiabatically from an initial temperature 127°C to a final temperature 27°C, decrease of internal energy of the mixture is: (take R=2 cal/mol-K)
detailed solution
Correct option is C
Decrease in internal energy =-∆U=(n1CV1+n2CV2+n3CV3)(T1-T2) =6×52R+4×32R+2×3R100=2700 R=5400 calTalk to our academic expert!
Similar Questions
100 mol of an ideal gas is heated from 10° to 20°C keeping its (i) volume constant (ii) pressure constant.
Let and denote the change in the internal energy of the gas due to process (i) and (ii),
respectively. Then, which of the following shall be true?
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
