Obtaining gamma (γ) in an adiabatic, reversible expansion

[anisotropic]

PROBLEM:

One mole of an ideal monatomic gas initially at 300 K (T₁) and a pressure of 15.0 atm (P₁) expands to a final pressure of 1.00 atm (P₂). The expansion occurs via an adiabatic and reversible path. Calculate q, w, ΔU, and ΔH.


SOLUTION:

q = 0 (adiabatic; no heat exchange occurs)
Thus, ΔU = w = C_vΔT

ΔH = C_pΔT

ΔT = T₂ - T₁

Need T₂ to calculate values (not an isothermal expansion)...

Known equation for reversible adiabatic process: P₁V₁^γ = P₂V₂^{γ (γ = gamma)}

Using PV = nRT, substitute in and simplify to obtain expression for T₂...

T₂ = T₁(P₁/P₂)<sup>((1-γ)/γ)

*where problems arise...</sup>

γ = ?

The solutions manual says γ = 5/3, but I don't know how this value is obtained.

I do know that, as an ideal monatomic gas, the internal energy (U) of the gas (per mole?) is: 3/2RT (we assume only translations occur). But how is this related to C_v, C_p, and more importantly, γ?

 

[Saitama]

γ can be obtained if you know the degrees of freedom for an ideal mono atomic gas.

 

[Infinitum]

The solutions manual says γ = 5/3, but I don't know how this value is obtained.

I do know that, as an ideal monatomic gas, the internal energy (U) of the gas (per mole?) is: 3/2RT (we assume only translations occur). But how is this related to C_v, C_p, and more importantly, γ?

\gamma = \frac{C_P}{C_V} = \frac{f+2}{f}

Where f is the number of degrees of freedom of the gas.

Taking the monoatomic case, the number of degrees of freedom are 3. So, \frac{2+3}{3} = \frac{5}{3}

Note that the internal energy U of one mole monoatomic ideal gas is

\frac{3}{2}RT

But in general, for any ideal gas, internal energy is

\frac{1}{2}fnRT