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  • 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 intial pressure. The value of n is: Option: 1 128 Option: 2</stro

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 intial pressure. The value of n is:
Option: 1 128
Option: 2 \frac{1}{3^2}
Option: 3 326
Option: 4 32

Answers (1)

best_answer

Correct Answer: a)128

\left(\frac{V_{1}}{V_{2}}\right)=\left(\frac{\rho_{2}}{\rho_{1}}\right)

Because

\left(\rho \propto \frac{1}{v}\right)

\begin{array}{l} \quad P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma} \\\\ \Rightarrow \quad\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(\frac{P_{2}}{P_{1}}\right) \end{array}

\begin{array}{l} \left(\frac{\rho_{2}}{\rho_{1}}\right)^{\gamma}=\frac{P_{2}}{P_{1}} \\ \\ \Rightarrow( 32)^{\frac{7}{5}}=\frac{P_{2}}{P_{1}} \end{array}

So,

\frac{P_2}{P_1} = 128

Posted by

Deependra Verma

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