# Thermodynamics, use of Cv and Cp

• MattHorbacz
In summary, 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?If you want to calculate the heat transferred to or from the system in a constant volume process you useQ = \Delta U = \int_{T_1}^{T_2} C_V \ dTIf you want to calculate the heat transferred to or from the

#### 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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For any adiabatic process Q = 0.

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

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
MexChemE said:
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?

MattHorbacz said:
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
MexChemE said:
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?

MattHorbacz said:
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.

MexChemE said:
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!

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##.

Juanda and MexChemE

## 1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships between heat, energy, and work. It is concerned with how energy is transferred between systems and how it can be converted from one form to another.

## 2. What is the difference between Cv and Cp?

Cv and Cp are both thermodynamic properties known as heat capacities. Cv (specific heat at constant volume) is the amount of heat required to raise the temperature of a substance by 1 degree at constant volume. Cp (specific heat at constant pressure) is the amount of heat required to raise the temperature of a substance by 1 degree at constant pressure. Cp is typically greater than Cv because it takes into account the work done by the substance as it expands against its surroundings.

## 3. How do you use Cv and Cp in thermodynamics?

Cv and Cp are used to calculate the amount of heat needed to change the temperature of a substance, as well as the amount of work done by the substance. They are also used to determine the change in internal energy and enthalpy of a system.

## 4. What are some real-world applications of thermodynamics?

Thermodynamics has many practical applications, including refrigeration and air conditioning, power generation, and engines. It is also important in understanding chemical reactions and phase changes in materials.

## 5. What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that the total energy of a closed system remains constant.