A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume in two separate experiments.The expansion is carried out at 300 K and at 600 K respectively.Choose the correct option.

Options

A. Work done at 600 K is 20 times the work done at 300 K
B. Work done at 300 K is twice the work done at 600 K
C. Work done at 600 K is twice the work done at 300 K
D. `DeltaU = 0` in both cases

Answer

To solve the problem of isothermal expansion of an ideal gas, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Conditions**:
We have 1 mole of an ideal gas that is expanded isothermally and reversibly to ten times its original volume at two different temperatures: 300 K and 600 K.

2. **Identify Key Relationships**:
- For an isothermal process, the change in internal energy (ΔU) is zero because ΔU = nCvΔT and ΔT = 0.
- The change in enthalpy (ΔH) is also zero for an isothermal process.

3. **Work Done in Isothermal Expansion**:
The work done (W) during an isothermal expansion of an ideal gas can be calculated using the formula:
\[ W = -2.303 \, nRT \, \log\left(\frac{V_f}{V_i}\right) \]
where:
- \( n \) = number of moles (1 mole)
- \( R \) = ideal gas constant (8.314 J/K·mol)
- \( T \) = temperature (300 K or 600 K)
- \( V_f \) = final volume (10 times the initial volume, \( V_i \))

4. **Calculate Work Done at 300 K**:
\[ W_1 = -2.303 \times 1 \times 8.314 \times 300 \times \log(10) \]
\[ W_1 = -2.303 \times 8.314 \times 300 \times 1 \]
\[ W_1 = -2.303 \times 2494.2 \]
\[ W_1 \approx -5742.5 \, J \]

5. **Calculate Work Done at 600 K**:
\[ W_2 = -2.303 \times 1 \times 8.314 \times 600 \times \log(10) \]
\[ W_2 = -2.303 \times 8.314 \times 600 \times 1 \]
\[ W_2 = -2.303 \times 4988.4 \]
\[ W_2 \approx -11485.0 \, J \]

6. **Compare Work Done**:
Since the work done is directly proportional to the temperature, we can express the relationship:
\[ \frac{W_2}{W_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2 \]
Thus, \( W_2 = 2 \times W_1 \).

7. **Conclusion**:
- The work done at 600 K is twice the work done at 300 K.
- The change in internal energy (ΔU) is zero for both cases.

### Final Answer:
- The correct options are:
- Work done at 600 K is twice the work done at 300 K.
- ΔU is 0 in both cases.