JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2020
- Medium
- Numerical
In an adiabatic process, the density of a diatomic gas becomes \(32\) times its initial value. The final pressure of the gas is found to be \(n\) times the initial pressure. The value of \(n\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(128\)
Step-by-step Solution
Detailed explanation
In adiabatic process
\(PV ^{\gamma}= constant\)
\(P \left(\frac{ m }{\rho}\right)^{\gamma}= constant\)
as mass is constant
\(P \propto \rho^{\gamma}\)
\(\frac{P_{f}}{P_{i}}=\left(\frac{\rho_{f}}{\rho_{i}}\right)^{\gamma}=(32)^{7 / 5}=2^{7}=128\)
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