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In previous, Find the value of gamma = C...

In previous, Find the value of `gamma = C_(P) // C_(v) :`

A

1.6

B

1.5

C

1.6

D

1.66

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The correct Answer is:
To find the value of \( \gamma = \frac{C_P}{C_V} \) for a gaseous mixture containing 2.0 moles of oxygen (O2) and 3.0 moles of carbon dioxide (CO2), we will follow these steps: ### Step 1: Identify the values of \( C_P \) and \( C_V \) for each gas - For oxygen (O2), \( C_P = 29 \, \text{J/(mol·K)} \) and \( C_V = 21 \, \text{J/(mol·K)} \). - For carbon dioxide (CO2), \( C_P = 37 \, \text{J/(mol·K)} \) and \( C_V = 28 \, \text{J/(mol·K)} \). ### Step 2: Calculate \( \gamma \) for each gas - For oxygen (O2): \[ \gamma_{O2} = \frac{C_{P_{O2}}}{C_{V_{O2}}} = \frac{29}{21} \approx 1.38 \] - For carbon dioxide (CO2): \[ \gamma_{CO2} = \frac{C_{P_{CO2}}}{C_{V_{CO2}}} = \frac{37}{28} \approx 1.32 \] ### Step 3: Calculate the average \( \gamma \) for the mixture To find the average \( \gamma \) for the mixture, we can use the mole fractions of each gas: - Total moles \( n_{total} = 2 + 3 = 5 \) - Mole fraction of O2: \[ x_{O2} = \frac{2}{5} = 0.4 \] - Mole fraction of CO2: \[ x_{CO2} = \frac{3}{5} = 0.6 \] Now, we can calculate the average \( \gamma \): \[ \gamma_{mix} = x_{O2} \cdot \gamma_{O2} + x_{CO2} \cdot \gamma_{CO2} \] Substituting the values: \[ \gamma_{mix} = 0.4 \cdot 1.38 + 0.6 \cdot 1.32 \] Calculating: \[ \gamma_{mix} = 0.552 + 0.792 = 1.344 \] ### Final Answer The value of \( \gamma \) for the gaseous mixture is approximately \( 1.344 \). ---

To find the value of \( \gamma = \frac{C_P}{C_V} \) for a gaseous mixture containing 2.0 moles of oxygen (O2) and 3.0 moles of carbon dioxide (CO2), we will follow these steps: ### Step 1: Identify the values of \( C_P \) and \( C_V \) for each gas - For oxygen (O2), \( C_P = 29 \, \text{J/(mol·K)} \) and \( C_V = 21 \, \text{J/(mol·K)} \). - For carbon dioxide (CO2), \( C_P = 37 \, \text{J/(mol·K)} \) and \( C_V = 28 \, \text{J/(mol·K)} \). ### Step 2: Calculate \( \gamma \) for each gas - For oxygen (O2): ...
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