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Step-by-step Solution
Step 1: Identify the given data
We have two different ideal gases mixed together:
• Gas 1: 2 moles with $ \gamma_1 = \frac{C_{p1}}{C_{v1}} = \frac{5}{3} $.
• Gas 2: 3 moles with $ \gamma_2 = \frac{C_{p2}}{C_{v2}} = \frac{4}{3} $.
Our goal is to find $ \gamma_{\text{mix}} = \frac{C_p}{C_v} $ for the mixture.
Step 2: Relate $C_p$ and $C_v$ to $\gamma$ for each gas
For an ideal gas, $ C_p - C_v = R $. Also, $ \gamma = \frac{C_p}{C_v} $. Hence, for each gas:
1. Gas 1:
$ \gamma_1 = \frac{C_{p1}}{C_{v1}} = \frac{5}{3}. $
This implies $ C_{p1} = \frac{5}{3} C_{v1} $. Since $ C_{p1} - C_{v1} = R $, we have:
$ \frac{5}{3}C_{v1} - C_{v1} = R \quad \Rightarrow \quad \frac{2}{3}C_{v1} = R \quad \Rightarrow \quad C_{v1} = \frac{3}{2}R. $
Thus $ C_{p1} = \frac{5}{2}R. $
2. Gas 2:
$ \gamma_2 = \frac{C_{p2}}{C_{v2}} = \frac{4}{3}. $
This implies $ C_{p2} = \frac{4}{3} C_{v2} $. Since $ C_{p2} - C_{v2} = R $, we get:
$ \frac{4}{3}C_{v2} - C_{v2} = R \quad \Rightarrow \quad \frac{1}{3}C_{v2} = R \quad \Rightarrow \quad C_{v2} = 3R. $
Hence $ C_{p2} = 4R. $
Step 3: Compute the effective $C_p$ of the mixture
If $n_1$ and $n_2$ are the moles of the two gases, then the effective $C_p$ is given by:
$$
C_p = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2}.
$$
Substituting $n_1 = 2$, $n_2 = 3$, $C_{p1} = \frac{5}{2} R$, and $C_{p2} = 4R$:
$$
C_p = \frac{2 \times \left(\frac{5}{2} R\right) + 3 \times (4R)}{2 + 3}
= \frac{5R + 12R}{5}
= \frac{17R}{5}
= 3.4R.
$$
Step 4: Compute the effective $C_v$ of the mixture
Similarly, the effective $C_v$ is:
$$
C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}.
$$
Substituting $n_1 = 2$, $n_2 = 3$, $C_{v1} = \frac{3}{2} R$, and $C_{v2} = 3R$:
$$
C_v = \frac{2 \times \left(\frac{3}{2} R\right) + 3 \times (3R)}{2 + 3}
= \frac{3R + 9R}{5}
= \frac{12R}{5}
= 2.4R.
$$
Step 5: Evaluate $ \gamma_{\text{mix}} $
Finally, the ratio $ \gamma_{\text{mix}} $ is:
$$
\gamma_{\text{mix}} = \frac{C_p}{C_v}
= \frac{3.4R}{2.4R}
= 1.4167 \approx 1.42.
$$
Hence, the value of $ \gamma $ for the mixture is 1.42.
