Question

if 16g of He is mixed with 32 g of O2 then calculate heat capacity ratio of the mixture

Answer 1

Given:

The mass of He = 16 g

The mass of O₂ = 32 g

To find:

The heat capacity ratio of the mixture

Solution:

Finding the no. of moles of bothe the gases:

The molar mass of He = 4 g/mol

The molar mass of O₂ = 32 g/mol

Let "n1" be the no. of moles of He and "n2" be the no. of moles of O₂

∴ n1 = $\frac{mass\: of\: He}{Molar\:Mass \:of\:He} = \frac{16}{4} = 4$

and

∴ n2 = $\frac{mass\: of\: O_2}{Molar\:Mass \:of\:O_2} = \frac{32}{32} = 1$

Finding the "$C_v$" specific heat capacity at constant volume and "$C_p$" specific heat capacity at constant pressure for the mixture:

Since

Helium is a mono-atomic gas ∴ γ₁ = $\frac{5}{3}$

Oxygen is a diatomic gas ∴ γ₂ = $\frac{7}{5}$

So,

The specific heat capacity at constant volume of He is,

$C_v_1$ = [R]/[γ₁ - 1] = [R] / [$\frac{5}{3} - 1$] = [R] / [$\frac{2}{3}$] = $\frac{3R}{2}$

The specific heat capacity at constant volume of O₂ is,

$C_v_2$ = [R]/[γ₂ - 1] = [R] / [$\frac{7}{5} - 1$] = [R] / [$\frac{2}{5}$] = $\frac{5R}{2}$

The specific heat capacity of the mixture at constant volume is given as,

$C_v = \frac{n_1C_V_1\:+\:n_2C_v_2}{n_1\:+\:n_2} = \frac{4*\frac{3R}{2} \:+\:1*\frac{5R}{2}}{4\:+\:1} = \frac{6R \:+\:\frac{5R}{2}}{5} = \frac{12R\:+\:5R}{10} = \frac{17R}{10}$

The specific heat capacity of the mixture at constant pressure is given as,

$C_p=C_v\:+\:R = \frac{17R}{10} \:+\:R = \frac{17R\:+\:10R}{10} = \frac{27R}{10}$

Finding the ratio of the specific heat capacity of the mixture:

∴ The ratio of the specific heat capacity of the mixture,

γ_mix = $\frac{C_p}{C_v}$ = $\frac{\frac{27R}{10}}{\frac{17R}{10} } = \frac{27}{17}$

Thus, the heat capacity ratio of the mixture is 27:17.

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Answer 2

The heat capacity ratio of the mixture is 27 : 17

Explanation:

The number of moles is given by the formula:

n = (Mass)/(Molar mass)

Now, the number of moles of helium:

n₁ = 16/4 = 4 moles

The number of moles of oxygen:

n₂ = 32/32 = 1 mole

The specific heat  at constant volume is given by the formula:

Cv = R/(γ - 1)

The specific heat of helium at constant volume :

Cv₁ = R/(5/3 - 1) = 3R/2

The specific heat of oxygen at constant volume :

Cv₂ = R/(7/5 - 1) = 5R/2

The specific heat of mixture at constant volume is given as:

Cv = ((n₁ × Cv₁) + (n₂ × Cv₂))/(n₁ + n₂)

On substituting the values, we get,

Cv = ((4 × 3R/2) + (1 × 5R/2))/(4 + 1)

∴ Cv = 17R/10

The specific heat of mixture at constant pressure is given as:

Cp = Cv + R = 17R/10 + R

∴ Cp = 27R/10

The ratio of specific heat capacity of the mixture is given as:

γ = Cp/Cv

On substituting the values, we get,

γ = (17R/10)/(27R/10)

∴ γ = 27/17 = 27 : 17