# Obtaining gamma (γ) in an adiabatic, reversible expansion

1. May 22, 2012

### 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 = CvΔT

ΔH = CpΔT

ΔT = T2 - T1

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

Known equation for reversible adiabatic process: P1V1γ = P2V2γ (γ = gamma)

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

T2 = T1(P1/P2)((1-γ)/γ)

*where problems arise...

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 Cv, Cp, and more importantly, γ?

Last edited: May 22, 2012
2. May 23, 2012

### Saitama

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

3. May 23, 2012

### Infinitum

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

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

$\frac{1}{2}fnRT$