An ideal mono-atomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. Calculate the ratio of final pressure to its initial pressure.
Show Hint
In adiabatic processes, the relationship between pressure and temperature for an ideal gas is governed by the adiabatic index \( \gamma \).
Step 1: Use the adiabatic condition.
For an adiabatic process, the relation between pressure and temperature is given by:
\[
P_1 T_1^{\gamma} = P_2 T_2^{\gamma}
\]
where \( \gamma \) is the adiabatic index, \( T_1 \) and \( T_2 \) are the initial and final temperatures, and \( P_1 \) and \( P_2 \) are the initial and final pressures.
Step 2: Apply the given condition.
It is given that the final temperature \( T_2 \) is twice the initial temperature \( T_1 \):
\[
T_2 = 2 T_1
\]
Substitute into the adiabatic equation:
\[
P_1 T_1^{\gamma} = P_2 (2 T_1)^{\gamma}
\]
\[
P_1 = P_2 \cdot 2^{\gamma}
\]
Thus, the ratio of final pressure to initial pressure is:
\[
\frac{P_2}{P_1} = 2^{\gamma}
\]
For a mono-atomic ideal gas, \( \gamma = \frac{5}{3} \), so:
\[
\frac{P_2}{P_1} = 2^{\frac{5}{3}} \approx 16
\]
