Intial temperature of a gas is 300 K & its `gamma` = 1.5 if adibatically its temperature reduced to 100 K, what will be effect on its volume & pressure.

Answer

To solve the problem step-by-step, we will analyze the effects on volume and pressure of a gas undergoing an adiabatic process, given the initial and final temperatures and the value of gamma.

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. The relationship between pressure (P), volume (V), and temperature (T) can be expressed using the following equations:
1. \( PV^{\gamma} = \text{constant} \)
2. \( TV^{\gamma - 1} = \text{constant} \)

### Step 2: Identify Given Values
- Initial temperature, \( T_1 = 300 \, K \)
- Final temperature, \( T_2 = 100 \, K \)
- Adiabatic constant, \( \gamma = 1.5 \)

### Step 3: Use the Temperature-Volume Relationship
Using the equation \( TV^{\gamma - 1} = \text{constant} \), we can write:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Substituting the known values:
\[ 300 \, V_1^{0.5} = 100 \, V_2^{0.5} \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ \frac{V_2^{0.5}}{V_1^{0.5}} = \frac{300}{100} = 3 \]
Squaring both sides:
\[ \frac{V_2}{V_1} = 9 \]
Thus, we find:
\[ V_2 = 9V_1 \]
This means the volume increases by a factor of 9.

### Step 5: Analyze the Effect on Pressure
Now, we will analyze the effect on pressure using the equation \( PV^{\gamma} = \text{constant} \):
\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]
Substituting \( V_2 = 9V_1 \):
\[ P_1 V_1^{1.5} = P_2 (9V_1)^{1.5} \]
This simplifies to:
\[ P_1 V_1^{1.5} = P_2 \cdot 27 V_1^{1.5} \]

### Step 6: Canceling Common Terms
Dividing both sides by \( V_1^{1.5} \):
\[ P_1 = 27 P_2 \]
Thus, we find:
\[ P_2 = \frac{P_1}{27} \]
This indicates that the pressure decreases to \( \frac{1}{27} \) of its initial value.

### Conclusion
- The volume of the gas increases to \( 9V_1 \).
- The pressure of the gas decreases to \( \frac{P_1}{27} \).