A given mass of a gas at `0^(@)C` is compressed reversible and adiabatically to a pressure 20 times the initial value. Calculate the final temperature of the gas. `[(C_(P))/(C_(V))=1.42`

Explanation

To solve the problem of finding the final temperature of a gas that is compressed adiabatically and reversibly, we can follow these steps:

### Step 1: Understand the given data
- Initial temperature, \( T_1 = 0^\circ C = 273.15 \, K \)
- The gas is compressed to a pressure \( P_2 = 20 P_1 \)
- The ratio of specific heats, \( \frac{C_p}{C_v} = \gamma = 1.42 \)

### Step 2: Use the adiabatic process equation
For an adiabatic process, the following relation holds:
\[ P_1^{\gamma} T_1^{\gamma - 1} = P_2^{\gamma} T_2^{\gamma - 1} \]

### Step 3: Substitute the known values
We know that \( P_2 = 20 P_1 \). We can substitute this into the equation:
\[ P_1^{\gamma} T_1^{\gamma - 1} = (20 P_1)^{\gamma} T_2^{\gamma - 1} \]

### Step 4: Simplify the equation
Dividing both sides by \( P_1^{\gamma} \):
\[ T_1^{\gamma - 1} = 20^{\gamma} T_2^{\gamma - 1} \]

### Step 5: Rearrange to find \( T_2 \)
Rearranging gives:
\[ T_2^{\gamma - 1} = \frac{T_1^{\gamma - 1}}{20^{\gamma}} \]
Taking the \( \frac{1}{\gamma - 1} \) power of both sides:
\[ T_2 = \left( \frac{T_1^{\gamma - 1}}{20^{\gamma}} \right)^{\frac{1}{\gamma - 1}} \]

### Step 6: Substitute the values
Substituting \( T_1 = 273.15 \, K \) and \( \gamma = 1.42 \):
\[ T_2 = \left( \frac{(273.15)^{1.42 - 1}}{20^{1.42}} \right)^{\frac{1}{1.42 - 1}} \]

### Step 7: Calculate \( T_2 \)
Calculating \( 1.42 - 1 = 0.42 \):
\[ T_2 = \left( \frac{(273.15)^{0.42}}{20^{1.42}} \right)^{\frac{1}{0.42}} \]

Calculating \( (273.15)^{0.42} \) and \( 20^{1.42} \):
1. \( (273.15)^{0.42} \approx 6.64 \)
2. \( 20^{1.42} \approx 39.81 \)

Now substituting these values:
\[ T_2 = \left( \frac{6.64}{39.81} \right)^{\frac{1}{0.42}} \]
Calculating \( \frac{6.64}{39.81} \approx 0.166 \):
\[ T_2 = (0.166)^{\frac{1}{0.42}} \approx 662.54 \, K \]

### Final Answer
The final temperature of the gas after adiabatic compression is approximately:
\[ T_2 \approx 662.54 \, K \]