Using Constant-Volume Specific Heat for Calculating Change of Internal Energy?

[KFC]

I note that in many case, we always use the constant-volume specific heat to calculate the change of internal energy.

For example, in a adiabatic process (P1, V1, T1) to (P2, V2, T2), since internal energy is state variable, we always like to build a fictitious isochoric process from (P1, V1, T1) to (P1', V1, T2) and isobaric process from (P1', V1, T2) to (P2, V2, T2) so that the total change of internal energy be

[tex]\Delta U = n C_v(T2-T1) + nC_p(T2-T2) = n C_v(T2-T1)[/tex]

Is this always true? There is a chapter about isobaric process in my text. The author use the constant-volume specific heat to calculate the change of internal energy

[tex]\Delta U = nC_v \Delta T[/tex]

the work done by the ideal gas is

[tex]\Delta W = nR\Delta T[/tex]

according to first law

[tex]\Delta Q = \Delta U + \Delta W = nC_v\Delta T + nR\Delta T = nC_p \Delta T[/tex]

this results is really confusing me. I wonder why don't we just use the constant-pressure specific heat to calculate the change of internal energy for isobaric process? But if we use [tex]C_p[/tex] to calculate [tex]\Delta U[/tex], the result will be different ... well all of these doubts is concluded in the following questions:

1) will it ALWAYS be true to use constant-volume specific heat to calculate the change of internal energy? No matter what process is concerned (even for isobaric process)?

2) The definition of heat capacity is: the change of heat per mole per degree. So why we keep use specific heat to calculate the change of internal energy instead of the change heat?

Thanks.

 

[Mapes]

1) will it ALWAYS be true to use constant-volume specific heat to calculate the change of internal energy? No matter what process is concerned (even for isobaric process)?

Only for an ideal gas, https://www.physicsforums.com/showpost.php?p=2009417&postcount=2". This is a result of the unique non-interaction of atoms in an ideal gas.

It might help to look at it this way: It's always true, for any system, that

[tex]dU=T\,dS-d(PV)+V\,dP[/tex]

For a constant-pressure process ([itex]dP=0[/itex]), the heat capacity

[tex]C_P=T\left(\frac{\partial S}{\partial T}\right)_P=C_V+nR[/tex]

for an ideal gas, and [itex]d(PV)=nR\,dT[/itex] for an ideal gas. So

[tex]\left(\frac{\partial U}{\partial T}\right)_P=C_V[/tex]

As you can see, [itex]dU=C_V dT[/itex] holds for constant-volume and constant-pressure processes. In fact it holds for all processes (for an ideal gas).

2) The definition of heat capacity is: the change of heat per mole per degree. So why we keep use specific heat to calculate the change of internal energy instead of the change heat?

The definition of heat capacity is [itex]T\left(\frac{\partial S}{\partial T}\right)_X[/itex], the heat needed to increase the temperature of a system by one degree under some condition X (there are also the molar heat capacity and specific heat capacity, which are normalized by amount of matter and mass, respectively).

 

[KFC]

Only for an ideal gas, https://www.physicsforums.com/showpost.php?p=2009417&postcount=2". This is a result of the unique non-interaction of atoms in an ideal gas.

It might help to look at it this way: It's always true, for any system, that

[tex]dU=T\,dS-d(PV)+V\,dP[/tex]

For a constant-pressure process ([itex]dP=0[/itex]), the heat capacity

[tex]C_P=T\left(\frac{\partial S}{\partial T}\right)_P=C_V+nR[/tex]

for an ideal gas, and [itex]d(PV)=nR\,dT[/itex] for an ideal gas. So

[tex]\left(\frac{\partial U}{\partial T}\right)_P=C_V[/tex]

As you can see, [itex]dU=C_V dT[/itex] holds for constant-volume and constant-pressure processes. In fact it holds for all processes (for an ideal gas).



The definition of heat capacity is [itex]T\left(\frac{\partial S}{\partial T}\right)_X[/itex], the heat needed to increase the temperature of a system by one degree under some condition X (there are also the molar heat capacity and specific heat capacity, which are normalized by amount of matter and mass, respectively).

Thanks again. It is really helpful.