Question:
If one mole of monoatomic gas (\(\gamma = \frac{5}{3}\)) is mixed with one mole diatomic gas (\(\gamma = \frac{7}{5}\)), the value of \(\gamma\) for the mixture is:
If one mole of monoatomic gas (\(\gamma = \frac{5}{3}\)) is mixed with one mole diatomic gas (\(\gamma = \frac{7}{5}\)), the value of \(\gamma\) for the mixture is:
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For mixtures, always add \(C_p\) and \(C_v\) separately.
Updated On: Apr 16, 2026
- \(1.40\)
- \(1.50\)
- \(1.53\)
- \(3.07\)
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The Correct Option is B
Solution and Explanation
Concept:
\[
\gamma = \frac{C_p}{C_v}
\]
Step 1: Use \(\gamma = \frac{C_p}{C_v}\).
Monoatomic: \[ C_v = \frac{3}{2}R,\quad C_p = \frac{5}{2}R \] Diatomic: \[ C_v = \frac{5}{2}R,\quad C_p = \frac{7}{2}R \]
Step 2: Total heat capacities.
\[ C_v^{total} = \frac{3}{2}R + \frac{5}{2}R = 4R \] \[ C_p^{total} = \frac{5}{2}R + \frac{7}{2}R = 6R \]
Step 3: Gamma of mixture.
\[ \gamma = \frac{6R}{4R} = 1.5 \]
Step 1: Use \(\gamma = \frac{C_p}{C_v}\).
Monoatomic: \[ C_v = \frac{3}{2}R,\quad C_p = \frac{5}{2}R \] Diatomic: \[ C_v = \frac{5}{2}R,\quad C_p = \frac{7}{2}R \]
Step 2: Total heat capacities.
\[ C_v^{total} = \frac{3}{2}R + \frac{5}{2}R = 4R \] \[ C_p^{total} = \frac{5}{2}R + \frac{7}{2}R = 6R \]
Step 3: Gamma of mixture.
\[ \gamma = \frac{6R}{4R} = 1.5 \]
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