1133a) 1 = 3 > 2 = 4 = 5 > 6
b) 1 = 3 = 5 > 6 > 2 = 4
c) 2 = 4 > 6 > 1 = 3 = 5
d) same as (b)
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1133a) 1 > 3 > 2 = 5 > 4 > 6
b) 5 = 3 = 1 > 6 > 4 > 2
c) 2 > 4 > 6 > 5 = 3 = 1
d) same as (b)
Akash Anand Some mistakes are there in all part..check it again.Upvote·0· Reply ·2014-02-15 02:05:39- Sourish Ghosh a) 1 = 3 > 2 = 4 = 5 > 6. Am I correct now?
- Akash Anand Yeah now its correct
1133
535(A) 1=3>2=4=5>6
(B) 1=3>2=4=5>6
(C) 2=4>6>1=3=5
(D) 1=3=5>6>2=4
- Akash Anand Check the second part again ...for the rest Excellent job :)
- Aditya Agarwal sir, isnt the increase in kinetic energy always equal to 3/2RdT.
whereas increase in U in Cv(dT).?
- Sourish Ghosh I have a doubt sir regarding this. What is the difference between 1/2fnRT and 3/2RT ?
- Akash Anand First one is the general formula for internal energy(U) of n mole of gas at any Temp.(T), where f is the degree of freedom.
Second one is the expression for internal energy(U) of 1 mole of a mono atomic gas at any Temp.(T)
- Sourish Ghosh Then the average kinetic energy of a gas = internal energy = 1/2fnRT ?
- Aditya Agarwal sir, i'm gettting the same result again.
in cases 1&3 = 30
for cases 2,4,and 5 = 18
for case 6 = 12.85
- Akash Anand @Sourish ..yeah..
- Akash Anand @Aditya..the given answer is:
1=3=5>6>2=4
481a) 1 = 3 > 2 = 4 = 5 > 6
c) 2 > 4 > 6 > 5 = 3 = 1
d)1 = 3 = 5 > 6 > 2 = 4
Sir how to do the second part,i am not understanding?
- Akash Anand Use this concept:
The average kinetic energy of a gas = internal energy = 1/2fnRT
481Sir But average K.E is given by 3/2kT,where k is boltzman constant,how do we get the relation average kinetic energy of a gas = internal energy = 1/2fnRT?
Niraj kumar Jha K.E. of a molecule of ideal monoatomic gas is 3/2kt.
K.E. of 'n' mole of gas with deg. of freedom 'f' is 1/2fnrt