106
Asish Mahapatra
·2009-02-10 02:27:56
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
·2009-02-10 02:28:53
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
106
Asish Mahapatra
·2009-02-10 08:24:56
well yea forgot the Rs was in hurry that time had full syllabus test
106
Asish Mahapatra
·2009-02-11 01:11:49
why degrees of freedom = 5
then Cv=5R/2 so Cp = Cv+R = 7R/2
or am i making a mistake