Question

The pressure and density of a diatomic gas (\( \gamma = 7/5 \)) change adiabatically from \( (p, d) \) to \( (p', d') \). If \( \frac{d'}{d} = 32 \), then \( \frac{p'}{p} \) should be:

• A. \( \frac{1}{128} \)
• B. \( 32 \)
• C. \( 128 \)
• D. \( \frac{1}{32} \)

Hint:

The adiabatic exponent \(\gamma = C_p/C_v\) determines the sensitivity of pressure changes to density changes during adiabatic compression or expansion.

Answer 1

The Correct Option is C

Step 1: Recall the adiabatic relation.
For an adiabatic change, $pV^\gamma = \text{const}$. Since density $d$ is inversely proportional to volume ($d \propto 1/V$), this rewrites neatly as $p \propto d^{\gamma}$.
Step 2: Turn it into a ratio.
Comparing the two states, \[ \frac{p'}{p} = \left(\frac{d'}{d}\right)^{\gamma}. \] This avoids needing any absolute values.
Step 3: Insert the given data.
Here $\gamma = \dfrac{7}{5}$ for a diatomic gas and $\dfrac{d'}{d} = 32$, so \[ \frac{p'}{p} = 32^{7/5}. \]
Step 4: Write 32 as a power of 2.
Since $32 = 2^5$, we get $32^{7/5} = (2^5)^{7/5}$. Writing it this way makes the fractional exponent cancel cleanly.
Step 5: Simplify the exponent.
$(2^5)^{7/5} = 2^{5\times 7/5} = 2^{7}$.
Step 6: Evaluate.
$2^7 = 128$, so $\dfrac{p'}{p} = 128$.
\[ \boxed{128} \]