A diatomic gas is first heated to double its volume at constant pressure and then heated at constant volume to double its pressure. Find the average molar heat capacity for this process.
Now do this for the reverse i.e. first it is heated at constant volume to double the pressure and then heated at constant pressure to double the volume.
Take degrees of freedom=5
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6 Answers
106for part (i)
Let initial temp. be T
PROCESS 1 - Isobaric
delta(Q1) = nCp*delta(T)
= n.7R/2.(2T-T)
= 7nRT/2
PROCESS 2 - Isochoric
delta(Q2) = nCv*delta(T)
= n.5R/2.(4T-2T)
= 10nRT/2
Average molar heat capacity = delta(Q=Q1+Q2)/[nR.delta(T)]
= 17nRT/2*n*3T
= 17R/6
106for part (ii)
Let initial temp. be T
PROCESS 1 - Isochoric
delta(Q1) = nCv*delta(T)
= n.5R/2.(2T-T)
= 5RnT/2
PROCESS 2 - Isobaric
delta(Q2) = nCp*delta(T)
= n.7R/2.(4T-2T)
= 14nRT/2
Average molar heat capacity = delta(Q=Q1+Q2)/[nR.delta(T)]
= 19nRT/2*n*3T
= 19R/6
106well yea forgot the Rs was in hurry that time had full syllabus test
106why degrees of freedom = 5
then Cv=5R/2 so Cp = Cv+R = 7R/2
or am i making a mistake