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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Feb 2008 00:09:26 IST
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1)Two moles of monoatomic gas is mixed with three moles of diatomic gas. The molar specific heat of the mixture at constant volume is ?? 2)If the density of a gas at ntp is 1.3mg/cc and velocity of sound in it is 330m/s. The number of degrees of freedom of gasmolecule is?? 3)A diatomic gas does 80J of work when expanded isobarically. The heat given to the gas during this process is ?? PLEASE HELP!!! (RATES FOR SURE!!!!)
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1)Two moles of monoatomic gas is mixed with three moles of diatomic gas. The molar specific heat of the mixture at constant volume is ??
 1 of monoatomic gas = (f+2)/f = (3+2)/3 = 5/3
 2 of diatomic gas = (5+2)/5 = 7/5
= [n1 1 + n22] / (n1+ n2)
= 1.506
Cv = R/ ( -1)
Cv = 16.43
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23 Feb 2008 00:36:59 IST
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2)If the density of a gas at ntp is 1.3mg/cc and velocity of sound in it is 330m/s. The number of degrees of freedom of gasmolecule is??
v =  (P/ )
330 =  (*1.013*105/1.3)
solving u get = 1.3975
use = (f+2)/f
u get 'f' = 5
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23 Feb 2008 00:38:29 IST
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I Understood the method, but the answer given is 2.1R (=17.451)
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23 Feb 2008 00:45:27 IST
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3)A diatomic gas does 80J of work when expanded isobarically. The heat given to the gas during this process is ??
W = PdV = nRdt
80 = ndTR
ndT = 9.62
dQ = dU +dW
nCpdT = nCvdt + dW
= (ndT)20.785 + 80 ................ using Cv=fR/2
= 9.62*20.785 + 80
= 279.95J
i hope its right
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23 Feb 2008 00:53:06 IST
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hey can you just explain me how ndt=9.2 ???
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23 Feb 2008 00:56:21 IST
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W = PdV
but PdV = nRdt ...............coz PV= nRT
so W =nRdT
u know the values of R(8.314) and W(80)
so u get the value of ndT
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23 Feb 2008 00:59:30 IST
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ok. i understood and could u pls check the first qn again!
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23 Feb 2008 06:48:35 IST
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i am telling u the shortcut formula--
f=degrees of freedom
f mixture=(n1f1 +n2f2 )/(n1+n2)
where n=number of moles
here f mixture=(2*3 +3*5)/(2+3)
=21/5
Cv=fR/2
=21R/10
=2.1R
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23 Feb 2008 10:20:45 IST
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nice method anchit
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23 Feb 2008 10:43:08 IST
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if u want to know the actual method, its this--
we use
dU=nCvdT
also
dU=nfRdT/2
therefore
we calculate the total internal energy of the gas--
dU=dT/2*(2*3*R+ 3*5*R)
=21R/2*dT
=nCvdT
where n=total number of moles=3+2=5
therefore
21R/2=5Cv
Cv=21R/10
=2.1R
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23 Feb 2008 12:52:20 IST
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THANKS FOR YOUR HELP ANCHIT!!!
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28 Feb 2008 01:16:13 IST
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do learn shorcuts but first understand how and from where it comes the given derivation is for anchit's shortcut
for mixture of two gases U = U1 + U2 ( U1 and U2 are the internal enargies of the mixed gases )
( f/2) nRT = (f1/2)n1RT + (f2/2)n2RT
f = ( n1f1 +n2f2)/ n1 + n2
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