Step 1: Understanding the Concept:
This problem compares different thermodynamic processes: isothermal, adiabatic, and isobaric.
We need to use the equation of state for each process to relate initial and final states.
Step 2: Key Formulas or Approach:
Isothermal process equation: \( P_i V_i = P_f V_f \).
Adiabatic process equation: \( P_i V_i^\gamma = P_f V_f^\gamma \).
Isobaric process equation: \( P = \text{constant} \).
Step 3: Detailed Explanation:
Let the initial volumes be \( V_i = V \) for all samples.
Let the final volume be \( V_f = 2V \) for all samples.
Let the final pressure be \( P_f = P \) for all samples, as given in the problem.
Let's find the initial pressure for each gas sample.
For sample X (Isothermal process):
Using the equation \( P_{X,i} V_i = P_{X,f} V_f \):
\[ P_{X,i} \cdot V = P \cdot (2V) \]
Solving for initial pressure:
\[ P_{X,i} = 2P \]
For sample Y (Adiabatic process):
Using the equation \( P_{Y,i} V_i^\gamma = P_{Y,f} V_f^\gamma \), with \( \gamma = \frac{3}{2} \):
\[ P_{Y,i} \cdot V^{3/2} = P \cdot (2V)^{3/2} \]
\[ P_{Y,i} \cdot V^{3/2} = P \cdot 2^{3/2} \cdot V^{3/2} \]
Cancel \( V^{3/2} \) from both sides:
\[ P_{Y,i} = P \cdot 2^{3/2} \]
Since \( 2^{3/2} = (\sqrt{2})^3 = 2\sqrt{2} \):
\[ P_{Y,i} = 2\sqrt{2}P \]
For sample Z (Isobaric process):
In an isobaric process, pressure remains constant throughout.
\[ P_{Z,i} = P_{Z,f} = P \]
Now, calculate the ratio of their initial pressures \( P_{X,i} : P_{Y,i} : P_{Z,i} \):
\[ \text{Ratio} = 2P : 2\sqrt{2}P : P \]
Divide by the common factor \( P \):
\[ \text{Ratio} = 2 : 2\sqrt{2} : 1 \]
Step 4: Final Answer:
The ratio of the initial pressures is \( 2 : 2\sqrt{2} : 1 \).