Reversible Adiabatic Expansion of a Perfect Mono-atomic Gas

In a reversible adiabatic process, there is no heat transfer between the system and its surroundings, and the process is carried out in a way that the system remains in thermodynamic equilibrium at all times. For a perfect mono-atomic gas, the relationship between its volume V and internal energy U during such an expansion can be determined using the first law of thermodynamics and the ideal gas law.

First Law of Thermodynamics:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

Since the process is adiabatic (Q = 0), the change in internal energy is solely due to the work done by the system:

ΔU = -W

Ideal Gas Law:
The ideal gas law relates the pressure, volume, and temperature of a gas:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

Reversible Adiabatic Process:
During a reversible adiabatic process, the relationship between pressure and volume for a perfect mono-atomic gas is given by:

P V^(γ) = constant

Where γ is the ratio of specific heat capacities for the gas, which is equal to 5/3 for a mono-atomic gas.

Derivation of the Relationship:
Using the ideal gas law, we can express the pressure in terms of the volume:

P = (nRT) / V

Substituting this into the equation for a reversible adiabatic process:

(nRT) / V V^(γ) = constant

(nRT) V^(γ-1) = constant

Since n, R, and T are constant for a given sample of gas, we can combine them into a single constant:

C = nRT

C V^(γ-1) = constant

Since the internal energy U of a perfect mono-atomic gas is only a function of its temperature, we can write U in terms of C:

U = f(C)

Taking the derivative of U with respect to V:

dU/dV = (dU/dC) (dC/dV)

Since U is only a function of C, the derivative dU/dC is a constant. And since C is constant during the expansion:

dU/dV = 0

This implies that U is constant during the reversible adiabatic expansion of a perfect mono-atomic gas.

Conclusion:
The relationship between volume V and internal energy U during a reversible adiabatic expansion of a perfect mono-atomic gas is given by:

UV^(2/3) = constant

Therefore, the correct option is (c) UV^(2/3) = constant.