Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas is container B is compressed to half of its original vlue adiabatically. The ratio of final pressure of gas of B to that of gas in A is

Options

A. `2^(gamma-1)`
B. `((1)/(2))^(gamma-1)`
C. `((1)/(1-gamma))^(2)`
D. `((1)/(gamma-1))^(2)`

Explanation

To solve the problem, we will analyze the two processes (isothermal and adiabatic) for the gases in containers A and B, respectively.

### Step 1: Understand the Initial Conditions
Both containers A and B contain identical gases at the same initial pressure (P₀), volume (V₀), and temperature (T₀).

### Step 2: Analyze the Isothermal Process in Container A
For container A, the gas is compressed isothermally to half its original volume. The final volume \( V_f \) is:
\[ V_f = \frac{V_0}{2} \]
Using the ideal gas law for isothermal processes, we have:
\[ P_i V_i = P_f V_f \]
Substituting the known values:
\[ P_0 V_0 = P_f \left(\frac{V_0}{2}\right) \]
Rearranging gives:
\[ P_f = \frac{P_0 V_0}{\frac{V_0}{2}} = 2 P_0 \]
Thus, the final pressure in container A is:
\[ P_f^A = 2 P_0 \]

### Step 3: Analyze the Adiabatic Process in Container B
For container B, the gas is compressed adiabatically to half its original volume. Again, the final volume \( V_f \) is:
\[ V_f = \frac{V_0}{2} \]
For adiabatic processes, the relation is given by:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
Substituting the known values:
\[ P_0 V_0^\gamma = P_f \left(\frac{V_0}{2}\right)^\gamma \]
Rearranging gives:
\[ P_f = P_0 \frac{V_0^\gamma}{\left(\frac{V_0}{2}\right)^\gamma} = P_0 \frac{V_0^\gamma}{\frac{V_0^\gamma}{2^\gamma}} = 2^\gamma P_0 \]
Thus, the final pressure in container B is:
\[ P_f^B = 2^\gamma P_0 \]

### Step 4: Calculate the Ratio of Final Pressures
We need to find the ratio of the final pressure of gas in container B to that in container A:
\[ \text{Ratio} = \frac{P_f^B}{P_f^A} = \frac{2^\gamma P_0}{2 P_0} \]
The \( P_0 \) cancels out:
\[ \text{Ratio} = \frac{2^\gamma}{2} = 2^{\gamma - 1} \]

### Final Answer
The ratio of the final pressure of gas in container B to that of gas in container A is:
\[ \frac{P_f^B}{P_f^A} = 2^{\gamma - 1} \]

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