For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible 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. `Delta U = 0` in both cases.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have a sample of 1.0 mol of an ideal gas that is expanded isothermally and reversibly to ten times its original volume. The expansion occurs at two different temperatures: 300 K and 600 K.

### Step 2: Write the Formula for Work Done
The work done (W) during the isothermal expansion of an ideal gas can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
where:
- \( n \) = number of moles (1.0 mol)
- \( R \) = universal gas constant (approximately 8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( V_f \) = final volume
- \( V_i \) = initial volume

Since the gas expands to ten times its original volume:
\[ V_f = 10V_i \]

### Step 3: Calculate Work Done at 300 K
Substituting the values for the first experiment at 300 K:
\[ W_{300} = - (1.0 \, \text{mol}) \cdot (8.314 \, \text{J/(mol·K)}) \cdot (300 \, \text{K}) \cdot \ln\left(\frac{10V_i}{V_i}\right) \]
\[ W_{300} = - (1.0) \cdot (8.314) \cdot (300) \cdot \ln(10) \]
\[ W_{300} = -2494.2 \cdot 2.3026 \]
\[ W_{300} \approx -5746.3 \, \text{J} \]

### Step 4: Calculate Work Done at 600 K
Now, substituting the values for the second experiment at 600 K:
\[ W_{600} = - (1.0 \, \text{mol}) \cdot (8.314 \, \text{J/(mol·K)}) \cdot (600 \, \text{K}) \cdot \ln\left(\frac{10V_i}{V_i}\right) \]
\[ W_{600} = - (1.0) \cdot (8.314) \cdot (600) \cdot \ln(10) \]
\[ W_{600} = - (1.0) \cdot (8.314) \cdot (600) \cdot 2.3026 \]
\[ W_{600} = - 12099.3 \, \text{J} \]

### Step 5: Compare Work Done at Both Temperatures
Now we can compare the work done at both temperatures:
\[ \frac{W_{300}}{W_{600}} = \frac{-5746.3}{-12099.3} \approx \frac{1}{2} \]
This indicates that:
\[ W_{600} = 2 \times W_{300} \]

### Step 6: Analyze the Change in Internal Energy (ΔU)
For an isothermal process involving an ideal gas, the change in internal energy (ΔU) is given by:
\[ \Delta U = nC_V(T_f - T_i) \]
Since the temperature is constant during the isothermal process, \( T_f = T_i \), thus:
\[ \Delta U = 0 \]

### Conclusion
From our calculations and analysis, we conclude:
- The work done at 600 K is twice that 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 zero in both cases.