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Thermal Physics

Blazing goIITian

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1 Mar 2009 21:13:10 IST

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Thermpdynamics

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JPS

Blazing goIITian

Joined: 21 Nov 2008

Posts: 1020

1 Mar 2009 21:16:51 IST

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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

saharsha kumar keshkar

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Joined: 7 Aug 2008

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2 Mar 2009 17:16:52 IST

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ans is (a)=2R...

if u want solution nudge me......

shraman asa

Blazing goIITian

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Posts: 611

2 Mar 2009 17:49:19 IST

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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 ...... :)

saharsha kumar keshkar

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Joined: 7 Aug 2008

Posts: 1104

2 Mar 2009 22:46:30 IST

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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 PV^x use the method...

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

where y is c_p/c_v

cheers!

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