A given mass of monoatomic gas occupies a volume of 4 litre at 1 atmosphere pressure and 300 K. It is compressed adiabatically to 1 litre. Find Final pressure and temperature

Answer

To solve the problem of finding the final pressure and temperature of a monoatomic gas that is compressed adiabatically from a volume of 4 liters to 1 liter, we will follow these steps:

### Step 1: Identify the given values
- Initial pressure \( P_1 = 1 \) atm
- Initial volume \( V_1 = 4 \) L
- Initial temperature \( T_1 = 300 \) K
- Final volume \( V_2 = 1 \) L

### Step 2: Determine the value of \( \gamma \) for a monoatomic gas
For a monoatomic gas, the degrees of freedom \( f = 3 \). The specific heats are defined as:
- \( C_V = \frac{3}{2} R \)
- \( C_P = C_V + R = \frac{5}{2} R \)

Thus, the ratio \( \gamma \) is given by:
\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{5}{2} R}{\frac{3}{2} R} = \frac{5}{3} \approx 1.67 \]

### Step 3: Use the adiabatic condition to find the final pressure \( P_2 \)
The adiabatic condition states that:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Rearranging gives:
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]
Substituting the known values:
\[ P_2 = 1 \, \text{atm} \left( \frac{4 \, \text{L}}{1 \, \text{L}} \right)^{1.67} \]
Calculating:
\[ P_2 = 1 \times 4^{1.67} \approx 10.13 \, \text{atm} \]

### Step 4: Use the relationship between temperature and volume to find the final temperature \( T_2 \)
The relationship for temperature in an adiabatic process is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Rearranging gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
Substituting the known values:
\[ T_2 = 300 \, \text{K} \left( \frac{4 \, \text{L}}{1 \, \text{L}} \right)^{1.67 - 1} \]
Calculating:
\[ T_2 = 300 \times 4^{0.67} \approx 300 \times 2.52 \approx 756 \, \text{K} \]

### Final Results
- Final Pressure \( P_2 \approx 10.13 \, \text{atm} \)
- Final Temperature \( T_2 \approx 756 \, \text{K} \)