Question

The pressure P₁ and density d₁ of diatomic gas \(\gamma = \frac{7}{5}\) changes suddenly to \(P_2 > P_1\) and d₂ respectively during an adiabatic process. The temperature of the gas increases and becomes _____ times of its initial temperature.
(Given \(\frac{d_2}{d_1} = 32\))

Answer 1

Correct Answer: 4

The process described is adiabatic, meaning there is no heat exchange. In such processes, the relation between pressure, volume, and temperature is given by:
\(P_1V_1^{\gamma} = P_2V_2^{\gamma}\). We also know  
\(PV = nRT\). Thus, \(P/(dR) = T\). For an adiabatic change, \(P \propto d^{\gamma}\) and \(P_1d_1^{-\gamma} = P_2d_2^{-\gamma}\). Given \(d_2/d_1 = 32\), substitute to get:

\(P_2/P_1 = (d_2/d_1)^{\gamma} = 32^{7/5}\).

Relating pressures and temperatures, \(T_2/T_1 = (P_2d_1)/(P_1d_2) = ((d_2/d_1)^{7/5}) \cdot (d_1/d_2)\).

Simplifying,\(T_2/T_1 = 32^{7/5 - 1} = 32^{2/5}\).

\(32^{2/5} = (2^5)^{2/5} = 2^2 = 4\).

The final temperature is 4 times the initial temperature, which falls within the range of 4,4.

Answer 2

\(P_1V_1^\gamma = P_2V_2^\gamma\)

\(\frac{P_1}{d_1^\gamma} = \frac{P_2}{d_2^\gamma}\)

\(\frac{d_1T_1}{d_1^\gamma} = \frac{d_2T_2}{d_2^\gamma}\)

\(T_2 = \left(\frac{d_2}{d_1}\right)^{\gamma-1} T_1\)

\(= (32)^{\frac{2}{5}} T_1\)
\(T_2 = 4T_1\)