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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

A

`2:1:sqrt2`

B

`2sqrt2:1:2`

C

`sqrt2:1:2`

D

`sqrt2:2:1`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the three different processes undergone by the gas samples A, B, and C, and derive the initial pressures based on the final pressures being equal. ### Step-by-Step Solution 1. **Identify Initial Conditions**: - Let the initial volume of each gas sample (A, B, C) be \( V \). - The final volume for each sample after doubling is \( V_2 = 2V \). - Let the initial pressures of samples A, B, and C be \( P_{1A}, P_{1B}, P_{1C} \). 2. **Final Pressures**: - Let the final pressure for all three samples be \( P \) (equal for all). 3. **Process for Sample A (Adiabatic)**: - The relationship for an adiabatic process is given by: \[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \] - Substituting the known values: \[ P_{1A} V^{\frac{3}{2}} = P (2V)^{\frac{3}{2}} \] - Simplifying: \[ P_{1A} V^{\frac{3}{2}} = P (2^{\frac{3}{2}} V^{\frac{3}{2}}) \] - Dividing both sides by \( V^{\frac{3}{2}} \): \[ P_{1A} = P \cdot 2^{\frac{3}{2}} = P \cdot 2\sqrt{2} \] 4. **Process for Sample B (Isobaric)**: - In an isobaric process, the pressure remains constant: \[ P_{1B} = P \] 5. **Process for Sample C (Isothermal)**: - The relationship for an isothermal process is given by: \[ P_1 V_1 = P_2 V_2 \] - Substituting the known values: \[ P_{1C} V = P (2V) \] - Simplifying: \[ P_{1C} = 2P \] 6. **Finding the Ratio of Initial Pressures**: - We now have: - \( P_{1A} = 2\sqrt{2} P \) - \( P_{1B} = P \) - \( P_{1C} = 2P \) - The ratio of the initial pressures \( P_{1A} : P_{1B} : P_{1C} \) is: \[ P_{1A} : P_{1B} : P_{1C} = 2\sqrt{2}P : P : 2P \] - Dividing each term by \( P \): \[ 2\sqrt{2} : 1 : 2 \] ### Final Answer The ratio of the initial pressures is: \[ 2\sqrt{2} : 1 : 2 \]
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