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 Increase in the internal energy

Answer

To find the increase in internal energy of a monoatomic gas that is compressed adiabatically from a volume of 4 liters to 1 liter, we can follow these steps:

### Step 1: Identify Given Values
- Initial volume \( V_1 = 4 \, \text{L} = 4 \times 10^{-3} \, \text{m}^3 \)
- Final volume \( V_2 = 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3 \)
- Initial pressure \( P_1 = 1 \, \text{atm} = 1.01325 \times 10^5 \, \text{Pa} \)
- Initial temperature \( T_1 = 300 \, \text{K} \)

### Step 2: Calculate the Number of Moles \( n \)
Using the ideal gas equation \( PV = nRT \):
\[ nR = \frac{P_1 V_1}{T_1} \]
Substituting the known values:
\[ nR = \frac{(1.01325 \times 10^5 \, \text{Pa}) \times (4 \times 10^{-3} \, \text{m}^3)}{300 \, \text{K}} = \frac{405.3}{300} = 1.351 \, \text{Pa m}^3/\text{K} \]

### Step 3: Determine the Final Temperature \( T_2 \)
For an adiabatic process, the relationship between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Where \( \gamma \) for a monoatomic gas is \( \frac{5}{3} \).
Rearranging gives:
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]
Substituting the values:
\[ T_2 = 300 \left( \frac{4}{1} \right)^{\frac{2}{3}} = 300 \times 4^{0.67} \approx 300 \times 2.52 \approx 756 \, \text{K} \]

### Step 4: Calculate Change in Temperature \( \Delta T \)
\[ \Delta T = T_2 - T_1 = 756 \, \text{K} - 300 \, \text{K} = 456 \, \text{K} \]

### Step 5: Calculate Change in Internal Energy \( \Delta U \)
For a monoatomic gas, the specific heat at constant volume \( C_V \) is given by:
\[ C_V = \frac{3}{2} R \]
Thus, the change in internal energy is:
\[ \Delta U = n C_V \Delta T = n \left( \frac{3}{2} R \right) \Delta T \]
Substituting \( nR = 1.351 \):
\[ \Delta U = n \left( \frac{3}{2} \times 1.351 \right) \times 456 \]
Calculating:
\[ \Delta U = 1.351 \times \frac{3}{2} \times 456 \approx 1.351 \times 2.25 \times 456 \approx 1350.5 \, \text{J} \]

### Final Answer
The increase in the internal energy of the gas is approximately:
\[ \Delta U \approx 930.16 \, \text{J} \]