Question

The pressure and density of a diatomic gas ($\gamma = \frac{7}{5}$) changes adiabatically from $(P, \rho)$ to $(P', \rho')$. If $\frac{\rho'}{\rho} = 32$ then $\frac{P'}{P}$ should be

• A. $\frac{1}{128}$
• B. 128
• C. 32
• D. 64

Hint:

Use the power of 2 for quick calculation: $32 = 2^5$, so $32^{7/5} = 2^7 = 128$.

Answer 1

The Correct Option is A

Step 1: For an adiabatic process, the relationship between pressure ($P$) and volume ($V$) is given by: \[ P V^{\gamma} = \text{constant} \]
Step 2: Since density ($\rho$) is mass per unit volume ($V = m/\rho$), volume is inversely proportional to density for a constant mass. Substituting $V \propto 1/\rho$ into the adiabatic equation: \[ P \left( \frac{1}{\rho} \right)^{\gamma} = \text{constant} \implies \frac{P}{\rho^{\gamma = \text{constant} \]
Step 3: Relate the initial and final states: \[ \frac{P'}{P} = \left( \frac{\rho'}{\rho} \right)^{\gamma} \]
Step 4: Substitute the given values $\frac{\rho'}{\rho} = 32$ and $\gamma = \frac{7}{5}$: \[ \frac{P'}{P} = (32)^{7/5} \] \[ \frac{P'}{P} = (2^5)^{7/5} = 2^7 = 128 \]