# Thermodynamics, use of Cv and Cp

MattHorbacz
I am studying for a thermo exam, and one of the problems I am doing deals with adiabatic expansion of a piston in a cylinder. When solving for work, the solution guide uses m*Cv*(T2-T1). I don't understand why they know how to use Cv instead of Cp. The pressure changes, so obviously you wouldn't use Cp, but isn't there also a change of volume since it is adiabatic expansion? It would seem that nether Cv nor Cv would be valid. What am I misinterpreting?

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## Answers and Replies

MexChemE
For any adiabatic process Q = 0.

MattHorbacz
For any adiabatic process Q = 0.
My mistake, I meant solving for Work, not heat

MexChemE
Oh, in that case, for an adiabatic process ΔU = W (Q + W notation), and for an ideal gas undergoing any kind of process this always holds:
$$\Delta U = C_V \Delta T$$
So the solution guide is correct, assuming your gas is ideal.

• Ravi Singh choudhary
MattHorbacz
Oh, in that case, for an adiabatic process ΔU = W (Q + W notation), and for an ideal gas undergoing any kind of process this always holds:
$$\Delta U = C_V \Delta T$$
So the solution guide is correct, assuming your gas is ideal.
Why can that be used even though volume isn't constant?

MexChemE
Why can that be used even though volume isn't constant?
Because U is a state function, and its change (ΔU) does not depend on the kind of process occurring. Furthermore, CV is defined as
$$C_V = \left( \frac{\partial U}{\partial T} \right)_V$$
And we know U only depends on temperature for ideal gases, so, for any ideal gas undergoing any kind of process, ΔU will always be
$$\Delta U = \int_{T_1}^{T_2} C_V \ dT$$
If CV were constant, this equation simplifies to the one I previously posted.

Now, this is what may be confusing you:
If you want to calculate the heat transferred to or from the system in a constant volume process you use
$$Q = \Delta U = \int_{T_1}^{T_2} C_V \ dT$$
If you want to calculate the heat transferred to or from the system in a constant pressure process you use
$$Q = \Delta H = \int_{T_1}^{T_2} C_P \ dT$$
But these equations apply only because Q = ΔU in a constant volume process, and Q = ΔH in a constant pressure process. However, for any kind of process, the definitions of ΔU and ΔH are always the same. That is why, in this case, we are having an adiabatic process, wherein neither pressure nor volume are constant, but the definition of ΔU is always the same, so you use CV regardless of the nature of the process.

I hope this helps clearing your doubts!

• Ravi Singh choudhary
MattHorbacz
Because U is a state function, and its change (ΔU) does not depend on the kind of process occurring. Furthermore, CV is defined as
$$C_V = \left( \frac{\partial U}{\partial T} \right)_V$$
And we know U only depends on temperature for ideal gases, so, for any ideal gas undergoing any kind of process, ΔU will always be
$$\Delta U = \int_{T_1}^{T_2} C_V \ dT$$
If CV were constant, this equation simplifies to the one I previously posted.

Now, this is what may be confusing you:
If you want to calculate the heat transferred to or from the system in a constant volume process you use
$$Q = \Delta U = \int_{T_1}^{T_2} C_V \ dT$$
If you want to calculate the heat transferred to or from the system in a constant pressure process you use
$$Q = \Delta H = \int_{T_1}^{T_2} C_P \ dT$$
But these equations apply only because Q = ΔU in a constant volume process, and Q = ΔH in a constant pressure process. However, for any kind of process, the definitions of ΔU and ΔH are always the same. That is why, in this case, we are having an adiabatic process, wherein neither pressure nor volume are constant, but the definition of ΔU is always the same, so you use CV regardless of the nature of the process.

I hope this helps clearing your doubts!

That makes more sense. But I still don't understand how you know to care about internal energy and not enthalpy. Is it just that for any case with ideal gasses consider internal energy?

MexChemE
That makes more sense. But I still don't understand how you know to care about internal energy and not enthalpy. Is it just that for any case with ideal gasses consider internal energy?
That is just a matter of convenience. When analyzing a thermodynamic system you always start from the First Law: ΔU = Q + W. In this adiabatic process, you need to calculate the work, so it can be done simply with W = ΔU, but this is not the only way to calculate the work in an adiabatic process.

Thermo is tough initially, but keep practicing, solve a lot of problems with a good textbook at hand and you will eventually get it.

MattHorbacz
That is just a matter of convenience. When analyzing a thermodynamic system you always start from the First Law: ΔU = Q + W. In this adiabatic process, you need to calculate the work, so it can be done simply with W = ΔU, but this is not the only way to calculate the work in an adiabatic process.

Thermo is tough initially, but keep practicing, solve a lot of problems with a good textbook at hand and you will eventually get it.

Ok thanks!

Mentor
We call ##C_v## the heat capacity at constant volume because that is how it can be measured experimentally, by measuring the amount of heat Q added at constant volume and dividing by the temperature change. But, as MexChemE pointed out, this physical property that we call ##C_v## has a more general meaning and applicability than that, ##C_v=(\partial U/\partial T)_V##. For an ideal gas, U(T,V) is a function only of T, and not V. The same, of course, goes for ##C_v##.

We call ##C_p## the heat capacity at constant pressure because that is how it can be measured experimentally, by measuring the amount of heat Q added at constant pressure and dividing by the temperature change. But, as MexChemE pointed out, this physical property that we call ##C_p## has a more general meaning and applicability than that, ##C_p=(\partial H/\partial T)_P##. For an ideal gas, H(T,P) is a function only of T, and not P. The same, of course, goes for ##C_p##.

• MexChemE