Question

Two ideal gases have the same value of C_p / C_v = γ. What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?

Answer 1

For the first ideal gas,
C_p1 = specific heat at constant pressure
C_v1 = specific heat at constant volume
n₁ = number of moles of the gas
$\frac{C_{{\mathrm{p}}_1}}{C_{{\mathrm{V}}_1}}$ = γ and C_p1 − C_v1 = R
⇒ γC_v1 − C_v1 = R
⇒ C_v1 (γ − 1) = R
$$\begin{gathered} \Rightarrow {\mathrm{Cv}}_1=\frac{R}{(\mathrm{\gamma }-1)} \\ C{\mathrm{p}}_1=\mathrm{\gamma }\frac{R}{(\mathrm{\gamma }-1)} \end{gathered}$$

For the second ideal gas,
C_p2 = specific heat at constant pressure
C_v2 = specific heat at constant volume
n₂ = number of moles of the gas
$\frac{{\mathrm{C}}_{{\mathrm{p}}_2}}{{\mathrm{C}}_{{\mathrm{v}}_2}}$ = γ and C_p2 − C_v2 = R
⇒ γC_v2 − C_v2 = R
⇒ C_v2 (γ − 1) = R
$$\begin{gathered} \Rightarrow {\mathrm{Cv}}_2=\frac{R}{(\mathrm{\gamma }-1)}\mathrm{and} \\ {\mathrm{Cp}}_2=\mathrm{\gamma }\frac{R}{(\mathrm{\gamma }-1)} \end{gathered}$$
Given:
n₁ = n₂ = 1 : 2
dU₁ = nC_v1dt
dU₂= 2nC_v2dT

When the gases are mixed,
nC_v1dT + 2nC_v2dT = 3nC_vdT
$$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\frac{{\mathrm{C}}_{v_1}+2{\mathrm{C}}_{v_2}}{3} \\ =\frac{{\displaystyle \frac{R}{(\mathrm{\gamma }-1)}}+{\displaystyle \frac{2R}{(\mathrm{\gamma }-1)}}}{3} \\ =\frac{3R}{(\mathrm{\gamma }-1)3}=\frac{R}{\mathrm{\gamma }-1} \end{gathered}$$

Hence, Cp / Cv in the mixture is γ.