Using the work-energy theorem in thermodynamics

If an $n$ moles of an ideal gas (one atomic) is heated at constant volume $V$ from temperature $T_1$ to 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 is identical.

Is my second solution reasoning valid?

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..