MHT CET · Physics · Thermodynamics
The pressure and density of a diatomic gas \(\left(\gamma=\frac{7}{5}\right)\) changes adiabatically from \((\mathrm{P}, \rho)\) to \(\left(\mathrm{P}^{\prime}, \rho^{\prime}\right)\). If \(\frac{\rho^{\prime}}{\rho}=32\) then \(\frac{\mathrm{P}^{\prime}}{\mathrm{P}}\) should be
- A \(\frac {1}{128}\)
- B \(128\)
- C \(32\)
- D \(64\)
Answer & Solution
Correct Answer
(B) \(128\)
Step-by-step Solution
Detailed explanation
For adiabatic process, \(\mathrm{PV}^\gamma=\mathrm{a}\) constant
\(\therefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\left(\frac{\mathrm{V}^{\prime}}{\mathrm{V}}\right)^\gamma=\left(\frac{\rho^{\prime}}{\rho}\right)^\gamma\)
Given that, \(\frac{\rho^{\prime}}{\rho}=32\) and \(\gamma=\frac{7}{5}\)
\(\therefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=(32)^{\frac{7}{5}}=128\)
\(\therefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\left(\frac{\mathrm{V}^{\prime}}{\mathrm{V}}\right)^\gamma=\left(\frac{\rho^{\prime}}{\rho}\right)^\gamma\)
Given that, \(\frac{\rho^{\prime}}{\rho}=32\) and \(\gamma=\frac{7}{5}\)
\(\therefore \quad \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=(32)^{\frac{7}{5}}=128\)
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