Question

Three perfect gases at absolute temperature T1,T2,T3 are mixed. The masses are m1,m2,m3 and no. of molecules are n1,n2,n3 resp. . Assuming no loss of energy , the final temperature of the mixture is?

Answer 1

i equated the internal energy of the mix to the sum of internal energies of these three.. but i am not able to eliminate Cv's at the end..how to do that?

Answer 2

@IrishBoy123 @samigupta8 @ParthKohli @imqwerty

Answer 3

@Vincent-Lyon.Fr

Answer 4

@ParthKohli

Answer 5

:)

Answer 6

>no loss of energy

Try using that? You know how to express energy in terms of moles and temperature.

Answer 7

yeah did you read abv. post of mine?

Answer 8

@zepdrix @jhonyy9 @johnweldon1993

Answer 9

\[(n_{1}+n_{2}+n_{3})Cv _{mix} T = n_{1}Cv_{1} T_{1} +n_{2}Cv_{2}T_{2}+n_{3}Cv_{3}T_{3}\]
How to eliminate the cv's from this eq. ?

Answer 10

*Cv 's (Heat capacity at const. volume)

Answer 11

So they've not assumed the same gas then? And does \(m_1\) mean mass or molar mass?

Answer 12

they have given they are 3 diff. gases..

Answer 13

and as for the masses i am not sure..i didn't use it at all..

Answer 14

@ganeshie8 @agent0smith

Answer 15

Can you use U=3/2 nRT?
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/shegas.html#c1

Answer 16

but they haven't mentioned whether they are monoatomic or diatomic or polyatomic..

Answer 17

@rsadhvika

Answer 18

@samigupta8

Answer 19

Assume they are same gases..

Answer 20

but they have given three perfect gases..not a gas is divided in three chambers..right?

Answer 21

@samigupta8

Answer 22

I meant the same atomicity gases..

Answer 23

Did you get it?

Answer 24

ok then Cv1=Cv2=Cv3 =3/2 R (i assumed they were monoatomic)...what will be Cv of the mixture? @samigupta8

Answer 25

3/2 R

Answer 26

how?

Answer 27

Formula for mixture Cv...

Answer 28

i know that Cv for a monatomic gas is 3/2 R but will it be the same for their mixture? how?

Answer 29

@samigupta8

Answer 30

Apply the formula for Cv mixture...!

Answer 31

oh yeah it will be the same only if they (gases) don't react with each other right?

Answer 32

Yep...

Answer 33

ok.. but still answer isn't matching..

Answer 34

you MUST be given more information. I think they've assumed the same nature of gas (mono/di-atomic)

Answer 35

yeah.. i assumed so.. but still answer isn't correct

Answer 36

what's the answer?

Answer 37

It is correct @priyar the fourth option ...
n1T1+n2T2+n3T3/n1+n2+n3

Answer 38

if i assume the nature to be same then their Cv's = 3/2 R hence get cancelled away..

Answer 39

is that the answer???

Answer 40

of course

Answer 41

i have been given with T1+T2+T3 in the denominator...!! only then i was wondering..

Answer 42

Thanks all!

Answer 43

Np...:)