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

${\displaystyle \frac{{\text{C}}_{\text{p}1}}{{\text{C}}_{\text{v}1}}=\gamma \text{and}{\text{C}}_{\text{p}1}-{\text{C}}_{\text{v}1}=\text{R}}$

${\displaystyle \Rightarrow \gamma {\text{C}}_{\text{v}1}-{\text{C}}_{\text{v}1}=\text{R}}$

${\displaystyle \Rightarrow {\text{C}}_{\text{v}1}(\gamma -1)=\text{R}}$

${\displaystyle \Rightarrow {\text{C}}_{\text{v}1}=\frac{\text{R}}{\gamma -1}}$

${\displaystyle {\text{C}}_{\text{p}1}=\gamma \frac{\text{R}}{(\gamma -1)}}$

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

${\displaystyle \frac{{\text{C}}_{\text{p}2}}{{\text{C}}_{\text{v}2}}=\gamma \text{and}{\text{C}}_{\text{p}2}-{\text{C}}_{\text{v}2}=\text{R}}$

${\displaystyle \Rightarrow \gamma {\text{C}}_{\text{v}2}-{\text{C}}_{\text{v}2}=\text{R}}$

${\displaystyle \Rightarrow {\text{C}}_{\text{v}2}(\gamma -1)=\text{R}}$

${\displaystyle \Rightarrow {\text{C}}_{\text{v}2}=\frac{\text{R}}{\gamma -1}}$

${\displaystyle {\text{C}}_{\text{p}2}=\gamma \frac{\text{R}}{(\gamma -1)}}$

Given:
n₁ = n₂ = 1 : 2
dU₁ = nC_v1dt
dU₂= 2nC_v2dT

When the gases are mixed,
nC_v1dT + 2nC_v2dT = 3nC_vdT

${\displaystyle {\text{C}}_{\text{v}}=\frac{{\text{C}}_{\text{v}1}+2{\text{C}}_{\text{v}2}}{3}}$

${\displaystyle =\frac{\frac{\text{R}}{\gamma -1}+\frac{2\text{R}}{\gamma -1}}{3}}$

${\displaystyle =\frac{3\text{R}}{(\gamma -1)3}=\frac{\text{R}}{\gamma -1}}$

Hence, Cp / Cv in the mixture is γ.