Step 1: Understanding the Concept:
An adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings.
For an ideal gas undergoing a reversible adiabatic process, the temperature and volume are related.
Step 2: Key Formula or Approach:
The relationship between temperature ($T$) and volume ($V$) for an adiabatic process is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
Therefore, for two states 1 (initial) and 2 (final):
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Step 3: Detailed Explanation:
Given values:
Adiabatic constant, $\gamma = \frac{7}{5}$
Initial Volume, $V_1 = V_0$
Final Volume, $V_2 = \frac{V_0}{32}$
Initial Temperature, $T_1 = T_i$
Final Temperature, $T_2 = xT_i$
Substitute these values into the adiabatic equation:
\[ T_i \cdot (V_0)^{\frac{7}{5} - 1} = (xT_i) \cdot \left(\frac{V_0}{32}\right)^{\frac{7}{5} - 1} \]
First, calculate the exponent: $\frac{7}{5} - 1 = \frac{7 - 5}{5} = \frac{2}{5}$.
The equation becomes:
\[ T_i \cdot V_0^{\frac{2}{5}} = xT_i \cdot \left(\frac{V_0}{32}\right)^{\frac{2}{5}} \]
We can cancel out the initial temperature $T_i$ from both sides:
\[ V_0^{\frac{2}{5}} = x \cdot \frac{V_0^{\frac{2}{5}}}{32^{\frac{2}{5}}} \]
Now, cancel out $V_0^{\frac{2}{5}}$ from both sides:
\[ 1 = x \cdot \frac{1}{32^{\frac{2}{5}}} \]
Solve for $x$:
\[ x = 32^{\frac{2}{5}} \]
To compute $32^{\frac{2}{5}}$, we can write it as $(32^{\frac{1}{5}})^2$.
Since $2^5 = 32$, the fifth root of 32 is 2 ($32^{\frac{1}{5}} = 2$).
\[ x = (2)^2 \]
\[ x = 4 \]
Step 4: Final Answer:
The value of $x$ is 4.