If $n$ moles of an ideal gas (one atomic) are heated at a constant volume $V$ from initial temperature $T_1$ to final temperature $T_2$, the amount of energy heat needed can be calculated as

$$ Q = nC_V \Delta T $$

Solving with work-energy theorem

By the work-energy theorem, the change in total kinetic energy is equal to the work done by net sum of the external forces ($\Delta K = W_f$). In this case, the work done on the gas is due to adding the heat $Q$ to the system. The total kinetic energy in a one atomic gas is given by $K = \frac{3}{2}pV = \frac{3}{2}nRT$. Therefore

$$ Q = \Delta K = K_2 - K_1 = \frac{3}{2}nR(T_2 - T_1) = \frac{3}{2}nR \Delta T$$

And since $C_V = \frac{3}{2}R$ for one atomic ideal gases, the two equations are identical.

Is my reasoning valid?


4 Answers 4


TL; DR: there is (mechanical) work and (thermodynamic) work.

Work-energy theorem

The work done by all forces acting on a particle (the work of the resultant force) equals the change in the kinetic energy of the particle.

is a theorem in classical mechanics, referring to the energy and work (done upon) of a point-like object.

In statistical mechanics we consider a very large collection of such objects (atoms or molecules) - of the order around Avogadro number of them. The work done on each atom/molecule obeys the work-energy theorem, but when we speak about the collection as a whole, we usually split this work into two components: (thermodynamic) work - corresponding to the macroscopic changes, and the work done on the microscopic level - which we call heat. The sum of the two correspond to the change of the total energy of the collection, and, in case of non-interacting particles (ideal gas) can be viewed as a manifestation of the work-energy theorem (for a non-ideal gas or a substance in another aggregate state, the total energy also includes the energy of interaction between the molecules, not covered by the work-energy theorem as formulate for one object.)


From the first law: ΔU = Q – W

Where: ΔU is the change in internal energy, Q is heat added to the gas, and W is work done by the gas which, for a closed system, is the integral of pdV

For a constant volume process (dV = 0), no work is done so W = 0. Moreover, for an ideal gas undergoing any process we have:

ΔU = n Cv ΔT

So Q in your example is simply heat added to the gas that results only in an increase in internal energy.


Let us try to understand the basics of First Law of Thermodynamics-

Heat (Q) and work (W) are the two ways to add or remove energy from a physical system.

n moles of an ideal gas (one atomic) is heated at constant volume V from temperature T1 to temperature T2, the amount of energy heat needed can be calculated as

Q = n C(v)ΔT

The processes are very different.

Heat is driven by temperature differences, while work involves a force exerted through a distance.

The heat and work can produce identical results.

For example, both can cause a temperature increase.

Heat transfers energy into a system, such as when the sun warms the air in a bicycle tire and increases the air’s temperature.

Similarly, work can be done on the system, as when someone pumps air into the tire. Heat and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy, U, of a system.

Internal energy is the sum of the kinetic and potential energies of a system’s atoms and molecules.

It can be divided into many subcategories, such as thermal and chemical energy, and depends only on the state of a system (that is, P, V, and T), not on how the energy enters or leaves the system. In order to understand the relationship between heat, work, and internal energy, we use the first law of thermodynamics. The first law of thermodynamics applies the conservation of energy principle to systems where heat and work are the methods of transferring energy into and out of the systems.

It can also be used to describe how energy transferred by heat is converted and transferred again by work.

so use ΔU = Q – W

and the process given by you says that no p.dv work is being done ,therefore the alternate calculation does not stand correct ,though it may be relating to change in temperature in the initial and final state of the system and trying to relate K.E. with work done, which is not happening..

Reference- https://www.texasgateway.org/resource/122-first-law-thermodynamics-thermal-energy-and-work


You are confusing the mechanical work with non-mechanical heat. Now, Joule did show that the two are 'interconvertible', albeit with some loss (the second law of thermodynamics), the two are not the same.

The first law states $dU = \partial W + \partial q$, where the signs may differ for different authors.

Correct statement: $\Delta U =\int \partial q$

Incorrect statement: $\int \partial W = \Delta K =\int \partial q$

$\Delta U = \int \partial W + \int \partial q$. For the isochoric case $dV = 0 \implies \int \partial W = 0$, we get $\Delta U =\int \partial q$. Nice though, that for a 'one atomic' (ideal) gas, the internal energy $U$ is equal to the sum of the kinetic energies of the gas particles ($K$), which, from Maxwell-Boltzmann distribution, for one mole, turns out to be $K = \dfrac{3}{2}RT$. So, numerically, for the isochoric case, your approach works, though it cannot be generalized for the not-isochoric cases.

You are actually obtaining the molar isochoric heat capacity $C_v$ using this approach. The molar isobaric heat capacity of ideal gases is $C_p = \dfrac{5}{2}RT$, giving the famous equation

$$ C_p-C_v = R $$


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