Question

5. Derive an expression for Cp and Cv for a
mixture of gases. Also, find the expressiɔn for
y of the mixture.​

Answer 1

Answer:

Hope it helps you........

Explanation:

Answer 2

Answer:

The expression for y of the mixture is $C_{p}=\frac{R Y}{\gamma-1}$

Explanation:

The molar heat capacity (C) of a substance is the total amount of energy in the form of heat required to raise the temperature of 1 mole of that substance by 1 unit. The nature, size, and composition of a substance in a system also play a big role.

$q=n C \Delta T$

For adiabatic process $\Delta Q=0 \Rightarrow \Delta U=-\Delta W$

$\Rightarrow \quad n C_{v} \Delta T=-\int P \cdot d V$

$\int_{v_{1}}^{v_{2}} p \cdot d V=\int_{v_{1}}^{v_{2}} k v^{-r} d V=k \int_{v_{1}}^{v_{2}} v^{-r} d V=\left.k \frac{v^{-r+1}}{1-\gamma}\right|_{v_{1}} ^{v_{2}}$

$\frac{k}{1-\gamma}\left[v_{2}^{1-\gamma}-v_{1}^{1-\gamma}\right]=\frac{R}{1-\gamma}\left[\left(k v_{2}^{-\gamma}\right) v_{2}-\left(k v_{1}^{-\gamma}\right) v_{1}\right]$

$\frac{P_{2} V_{2}-P_{1} V_{1}}{1-\gamma} \Rightarrow n C_{V} \Delta T=\left(\frac{\Delta P V}{1-\gamma}\right)(-1)$

we know that $P V=n R T$

$\Delta P V=\Delta n R T=n R \Delta T$

$\gamma=\frac{C_{p}}{C_{v}}$

The heat capacity ratio is used in reversible thermodynamic processes, especially when ideal gases are involved.

$n C V \Delta T=\frac{n R \Delta T}{r-1} \Rightarrow C V=\frac{R}{r-1}$

$\frac{C_{F}}{C_{v}}=r \Rightarrow C_{p}=\frac{R Y}{\gamma-1}$