A diatomic gas at STP is expanded to thirty two times its volume under adiabatic conditions the resulting temprature in kelvin ?(answer must be in two decimal points )

Options

A. Relative velocity of the fluid is geater
B. Relative velocity of the fluid is smaller
C. Pressure is greater
D. Kinetic energy of the fluid is smaaler

Answer

To solve the problem step by step, we will use the principles of thermodynamics related to adiabatic processes for a diatomic gas.

### Step 1: Identify the Initial Conditions
We know that the gas is at Standard Temperature and Pressure (STP). The initial temperature \( T_i \) is given as:
\[ T_i = 273 \text{ K} \]
The initial volume \( V_i \) can be assumed as \( V \).

### Step 2: Determine the Final Volume
The gas expands to 32 times its initial volume:
\[ V_f = 32 V_i = 32 V \]

### Step 3: Use the Adiabatic Condition
For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V^{\gamma - 1} = T_2 V^{\gamma - 1} \]
Where \( \gamma \) (gamma) for a diatomic gas is:
\[ \gamma = \frac{5}{2} \]

### Step 4: Substitute the Known Values
We can rewrite the equation using the initial and final conditions:
\[ T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1} \]
Substituting the known values:
\[ 273 \cdot V^{\frac{5}{2} - 1} = T_f \cdot (32V)^{\frac{5}{2} - 1} \]

### Step 5: Simplify the Equation
We can simplify the equation:
\[ 273 \cdot V^{\frac{3}{2}} = T_f \cdot (32^{\frac{3}{2}} \cdot V^{\frac{3}{2}}) \]
Dividing both sides by \( V^{\frac{3}{2}} \):
\[ 273 = T_f \cdot 32^{\frac{3}{2}} \]

### Step 6: Solve for Final Temperature \( T_f \)
Now, we can isolate \( T_f \):
\[ T_f = \frac{273}{32^{\frac{3}{2}}} \]

### Step 7: Calculate \( 32^{\frac{3}{2}} \)
Calculating \( 32^{\frac{3}{2}} \):
\[ 32^{\frac{3}{2}} = (32^{1/2})^3 = (4)^3 = 64 \]

### Step 8: Substitute Back to Find \( T_f \)
Now substitute back:
\[ T_f = \frac{273}{64} \]
Calculating \( T_f \):
\[ T_f = 4.265625 \text{ K} \]

### Step 9: Round to Two Decimal Points
Finally, rounding to two decimal points:
\[ T_f \approx 4.27 \text{ K} \]

### Final Answer
The resulting temperature in Kelvin is:
\[ \boxed{4.27 \text{ K}} \]