A monatomic gas is expanded adiabatically and due to expansion volume becomes 8 times, then

Options

A. a.Pressure becomes `(1)/(32)` times
B. b.Total K.E. of gas molecules becomes `(1)/(4)` times
C. c.Ratio of translational and roational kinetic energy changes with variation in temperature
D. d.None of These

Answer

To solve the problem step by step, we need to analyze the adiabatic expansion of a monatomic gas where the volume becomes 8 times its initial value. We will use the principles of thermodynamics, particularly the equations governing adiabatic processes.

### Step 1: Understand the adiabatic process
In an adiabatic process, there is no heat exchange with the surroundings. For a monatomic ideal gas, the relationship between pressure (P), volume (V), and temperature (T) is given by:
\[ PV^\gamma = \text{constant} \]
where \(\gamma\) (gamma) for a monatomic gas is \(\frac{5}{3}\).

### Step 2: Set up the initial and final states
Let:
- Initial pressure = \(P_1\)
- Initial volume = \(V_1\)
- Final pressure = \(P_2\)
- Final volume = \(V_2 = 8V_1\)

Using the adiabatic condition:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

### Step 3: Substitute the final volume
Substituting \(V_2 = 8V_1\) into the equation:
\[ P_1 V_1^{\frac{5}{3}} = P_2 (8V_1)^{\frac{5}{3}} \]

### Step 4: Simplify the equation
This simplifies to:
\[ P_1 V_1^{\frac{5}{3}} = P_2 \cdot 8^{\frac{5}{3}} V_1^{\frac{5}{3}} \]
Cancelling \(V_1^{\frac{5}{3}}\) from both sides gives:
\[ P_1 = P_2 \cdot 8^{\frac{5}{3}} \]

### Step 5: Calculate \(8^{\frac{5}{3}}\)
Calculating \(8^{\frac{5}{3}}\):
\[ 8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^5 = 32 \]

### Step 6: Relate the pressures
Thus, we have:
\[ P_1 = 32 P_2 \]
Rearranging gives:
\[ P_2 = \frac{P_1}{32} \]

### Step 7: Analyze the kinetic energy
The total kinetic energy of the gas is related to its temperature. The kinetic energy of a monatomic ideal gas is given by:
\[ KE = \frac{3}{2} nRT \]
Since \(PV = nRT\), we can express temperature in terms of pressure and volume:
\[ T = \frac{PV}{nR} \]

### Step 8: Find the temperature ratio
Using the initial and final states:
\[ \frac{T_2}{T_1} = \frac{P_2 V_2}{P_1 V_1} \]
Substituting \(P_2 = \frac{P_1}{32}\) and \(V_2 = 8V_1\):
\[ \frac{T_2}{T_1} = \frac{\left(\frac{P_1}{32}\right)(8V_1)}{P_1 V_1} = \frac{8}{32} = \frac{1}{4} \]

### Step 9: Relate kinetic energy to temperature
Since kinetic energy is proportional to temperature:
\[ KE_2 = \frac{3}{2} nR T_2 = \frac{3}{2} nR \left(\frac{1}{4} T_1\right) = \frac{1}{4} KE_1 \]

### Conclusion
From the analysis:
- The final pressure \(P_2\) is \(\frac{P_1}{32}\).
- The total kinetic energy of the gas molecules becomes \(\frac{1}{4}\) of the initial kinetic energy.

### Final Answers
1. Pressure becomes \(\frac{1}{32}\) times the initial pressure.
2. Kinetic energy of the gas molecules becomes \(\frac{1}{4}\) times the initial kinetic energy.
3. The ratio of translational and rotational kinetic energy does not change with temperature.