One mole of an ideal gas expands from state-I (Atm , 20 litre) to (2 Atm , 10 litre ) isothermally .
Caluclate , `w & DeltaU`.

Answer

To solve the problem of calculating the work done (w) and the change in internal energy (ΔU) for the isothermal expansion of one mole of an ideal gas, we can follow these steps:

### Step 1: Understand the conditions of the problem
- We have one mole of an ideal gas.
- The initial state (State I) is at a pressure of 1 atm and a volume of 20 liters.
- The final state (State II) is at a pressure of 2 atm and a volume of 10 liters.
- The process is isothermal, meaning the temperature remains constant.

### Step 2: Calculate the work done (w)
For an isothermal process, the work done by the gas during expansion can be calculated using the formula:
\[ w = -nRT \ln \left( \frac{V_2}{V_1} \right) \]
Where:
- \( n \) = number of moles (1 mole)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (we need to find this)
- \( V_1 \) = initial volume (20 L)
- \( V_2 \) = final volume (10 L)

### Step 3: Find the temperature (T)
Since we are not provided with the temperature directly, we can use the ideal gas law to find it at State I:
\[ PV = nRT \]
Rearranging gives:
\[ T = \frac{PV}{nR} \]
Substituting the values for State I:
- \( P = 1 \) atm
- \( V = 20 \) L
- \( n = 1 \) mole
- \( R = 0.0821 \) L·atm/(K·mol)

Calculating T:
\[ T = \frac{(1 \, \text{atm})(20 \, \text{L})}{(1 \, \text{mol})(0.0821 \, \text{L·atm/(K·mol)})} = \frac{20}{0.0821} \approx 243.5 \, K \]

### Step 4: Substitute values into the work formula
Now we can substitute the values into the work formula:
\[ w = -1 \times 0.0821 \times 243.5 \ln \left( \frac{10}{20} \right) \]
Calculating \( \ln \left( \frac{10}{20} \right) = \ln(0.5) \):
\[ \ln(0.5) \approx -0.693 \]
Now substituting:
\[ w = -1 \times 0.0821 \times 243.5 \times (-0.693) \]
Calculating:
\[ w \approx 0.0821 \times 243.5 \times 0.693 \approx 14.0 \, \text{L·atm} \]

### Step 5: Calculate the change in internal energy (ΔU)
For an isothermal process involving an ideal gas, the change in internal energy is given by:
\[ \Delta U = 0 \]
This is because the internal energy of an ideal gas depends only on temperature, and since the temperature is constant, the change in internal energy is zero.

### Final Answers:
- Work done (w) = 14.0 L·atm
- Change in internal energy (ΔU) = 0