10-02-08 Average Heat capacity.

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

6 Answers

106
Asish Mahapatra ·

for 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

106
Asish Mahapatra ·

for 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

62
Lokesh Verma ·

ashish Where is "R"

is Cp=7/2? :O

106
Asish Mahapatra ·

well yea forgot the Rs was in hurry that time had full syllabus test

106
Asish Mahapatra ·

why degrees of freedom = 5
then Cv=5R/2 so Cp = Cv+R = 7R/2

or am i making a mistake

62
Lokesh Verma ·

I went mad then :D

will delete that ;)

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