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[Post New] 1 Mar 2009 21:13:10 IST
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JPS (1695)

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[Post New] 1 Mar 2009 21:16:51 IST
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JPS (1695)

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Q. 37

P-V graph of an ideal monoatomic gas is as shown. Molar heat capacity of gas will be

 

      a.

2 R

 
 

      b.

3 R

 
 

      c.

5 R

 
 

      d.

7 R

 
 

 

 
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[Post New] 2 Mar 2009 17:16:52 IST
   Subject: Thermpdynamics
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saharsha kumar keshkar (3397)

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ans is (a)=2R...

if u want solution nudge me......
B Tech- IIIrd year
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IIT Bombay
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[Post New] 2 Mar 2009 17:49:19 IST
   Subject: Thermpdynamics
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shraman asa (713)

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nCdT = nCvdT + PdV ......

C = Cv + PdV/(n)dT....

frm graph P=kV ....

PV=nRT .....

dV/dT = nR/2kV ........

BUT P = kV ...........

THUS THE SECOND TERM ON RHS IS R/2 ....

SINCE ITS MONOATOMIC Cv = 3R/2 .....

thus R/2 + 3R/2 = 2R ......  :)
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[Post New] 2 Mar 2009 22:46:30 IST
   Subject: Thermpdynamics
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saharsha kumar keshkar (3397)

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 correct sharman !

thats the basic way to do the problem....

but any way take another method.....(formula)

for any process type PVx use the method...

 

C=(R/(y-1))+(R/(1-x))

where y is cp/cv

cheers!

 
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