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Step-by-Step Detailed Solution
Step 1: Use the Adiabatic Process Relation
In an adiabatic process for an ideal gas, the pressure $P$ and the volume $V$ are related by the equation:
$P\,V^\gamma = \text{constant},$
where $ \gamma = \frac{C_p}{C_v} $.
Step 2: Express the Pressure Ratio in Terms of Volume Ratio
Let the initial pressure be $P_i$ and initial volume be $V_i$, the final pressure be $P_f$ and the final volume be $V_f$. From the adiabatic relation:
$ \frac{P_i}{P_f} = \Bigl(\frac{V_f}{V_i}\Bigr)^\gamma. $
Step 3: Substitute the Given Values
Initial volume, $V_i = 0.8\,\text{L}$.
Final volume, $V_f = 2.7\,\text{L}$.
The ratio $ \frac{P_f}{P_i} = \frac{16}{81} $ ⇒ thus $ \frac{P_i}{P_f} = \frac{81}{16} $.
So the equation becomes:
$ \frac{81}{16} = \biggl(\frac{2.7}{0.8}\biggr)^\gamma. $
Step 4: Simplify the Volume Ratio
$ \frac{2.7}{0.8} = \frac{27}{8}. $
Hence the equation is now:
$ \frac{81}{16} = \Bigl(\frac{27}{8}\Bigr)^\gamma. $
Step 5: Convert to Exponential Forms and Solve for $\gamma$
$81 = 3^4$
$16 = 2^4$
$27 = 3^3$
$8 = 2^3$
So we rewrite:
$ \Bigl(\frac{3^3}{2^3}\Bigr)^\gamma = \frac{3^4}{2^4}. $
This implies:
$ 3^{3\gamma}\,2^{-3\gamma} = 3^4 \, 2^{-4}. $
Equating exponents on both sides for the base 3:
$ 3\gamma = 4 \quad \Rightarrow \quad \gamma = \frac{4}{3}.
This satisfies the exponent matching for the base 2 as well.
Step 6: Final Answer
Therefore, the ratio $ \displaystyle \frac{C_p}{C_v} $ is:
$ \gamma = \frac{4}{3}. $
