Three samples of the same gas A,B and C `(gamma=3//2)` have initially equal volume. Now the volume of each sample is doubled. The process is adiabatic for A. Isobaric for B and isothermal for C. If the final pressures are equal for all three samples, find the ratio of their initial pressures

Options

A. `2sqrt2:2:1`
B. `2sqrt2:1:2`
C. `sqrt2:1:2`
D. `2:1:sqrt2`

Answer

To solve the problem, we need to analyze the three different processes (adiabatic, isobaric, and isothermal) that the gas samples A, B, and C undergo when their volumes are doubled. We will find the initial pressures of each sample and then determine the ratio of these pressures.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
Let the initial volume of each gas sample be \( V \) and the initial pressure of samples A, B, and C be \( P_A, P_B, \) and \( P_C \) respectively.

2. **Final Conditions for Each Process:**
- For sample A (adiabatic process):
The relationship for an adiabatic process is given by:
\[ P_A V_A^{\gamma} = P_{A_f} V_{A_f}^{\gamma} \]
where \( V_{A_f} = 2V \) and \( \gamma = \frac{3}{2} \).
Thus, we have:
\[ P_A V^{\frac{3}{2}} = P_{A_f} (2V)^{\frac{3}{2}} \]
Simplifying gives:
\[ P_{A_f} = P_A \left(\frac{V}{2V}\right)^{\frac{3}{2}} = P_A \left(\frac{1}{2}\right)^{\frac{3}{2}} = P_A \cdot \frac{1}{2\sqrt{2}} \]

- For sample B (isobaric process):
The pressure remains constant, so:
\[ P_{B_f} = P_B \]

- For sample C (isothermal process):
The relationship for an isothermal process is given by:
\[ P_C V_C = P_{C_f} V_{C_f} \]
where \( V_{C_f} = 2V \). Thus:
\[ P_C V = P_{C_f} (2V) \]
Simplifying gives:
\[ P_{C_f} = \frac{P_C V}{2V} = \frac{P_C}{2} \]

3. **Set Final Pressures Equal:**
Since the final pressures are equal for all three samples, we can set them equal:
\[ P_{A_f} = P_{B_f} = P_{C_f} \]
Substituting the expressions we derived:
\[ P_A \cdot \frac{1}{2\sqrt{2}} = P_B = \frac{P_C}{2} \]

4. **Express Pressures in terms of a Common Variable:**
Let’s express \( P_B \) in terms of \( P_A \):
\[ P_B = P_A \cdot \frac{1}{2\sqrt{2}} \]
And express \( P_C \) in terms of \( P_B \):
\[ P_C = 2P_B = 2 \left(P_A \cdot \frac{1}{2\sqrt{2}}\right) = \frac{P_A}{\sqrt{2}} \]

5. **Ratio of Initial Pressures:**
Now we have:
\[ P_A : P_B : P_C = P_A : \left(P_A \cdot \frac{1}{2\sqrt{2}}\right) : \left(\frac{P_A}{\sqrt{2}}\right) \]
This simplifies to:
\[ 1 : \frac{1}{2\sqrt{2}} : \frac{1}{\sqrt{2}} \]
To eliminate the fractions, multiply through by \( 2\sqrt{2} \):
\[ 2\sqrt{2} : 1 : 2 \]
Thus, the ratio of the initial pressures is:
\[ 2\sqrt{2} : 1 : 2 \]

### Final Answer:
The ratio of the initial pressures \( P_A : P_B : P_C \) is \( 2\sqrt{2} : 1 : 2 \).