An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps ate Q1=5960J,Q2=−5585J,Q3=−2980J, Q4=3645J, respectively. The corresponding works involved are W1=2200J,W2=−825J,W3=−1100J and W4, respectively. The value of W4 is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
1315 J
b
275 J
c
765 J
d
675 J
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Q=Q1+Q2+Q3+Q4 =5960−5585−2980+3645=1040JW=W1+W2+W3+W4 =2200−825−1100+W4=275+W4 For a cyclic process, Uf=Uf ΔU=Uf−Ui=0From the first law of thermodynamics, Q=ΔU+W1040=0=275+W4 or W4=765J