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Dear Maam

Please explain me the derivation of Cp -Cv=R

Thank you

 

We know that at constant pressure work done by a gas during its expansion can be expressed as:

w = P∆V

Therefore, for one mole of an ideal gas


PV = RT ……………………….. (1)
 

When temperature is raised by 1oC from T to T + 1 so that volume becomes V + ∆V, then

P (V + ∆V) = R (T +1) ………………………………… (2)

Subtracting equation (1) from equation (2), we get:

P∆V = R

Thus, at constant pressure work done by one mole of the an ideal gas in expansion when heated through 1oC is equal to R

Hence, CP – CV = R

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 the derivation goes like this:

since q=C*deltaT,

qv=Cv*delta T and qp=Cp*delta T. also, qv=delta U, and qp=delta H;

therefore, Cv*delta T=delta U and Cp*delta T=delta H

now,fro 1 mole of an ideal gas

delta H=delta U+delta(pV)

=delta U + delta(RT) {from the equation pV=nRT}

delta U + R*delta T

substituing the values, we get

Cp*delta T=Cv*delta T + R*delta T

dividing by delta T, we get,

Cp=Cv +R 

hence Cp-Cv=R

hope that helps!

cheers!

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jayanthi cn u plz xplain d 4th line ? wats dus h reprsnt ?

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 Thanks

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As we know that at constant pressure work done by a gas in expansion can be expressed as:

w= P∆V

Therefore, for one mole of an ideal gas:


PV = RT ……………………….. (1)
 

When temperature is raised by 1oC from T to T + 1 so that volume becomes V + ∆V, then

P (V + ∆V) = R (T +1) ………………………………… (2)

Subtracting equation (1) from equation (2), we get:
 

P∆V = R

Thus, At constant pressure work done by one mole of the an ideal gas in expansion when heated through 1oC is equal to R

Hence, CP – CV = R

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