For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from a volume of `10 dm^(3)` to `20 dm^(3), Delta H` is -

Options

A. 1.73kJ
B. `-1.73kJ`
C. 3.46kJ
D. Zero

Answer

To find the enthalpy change (ΔH) for the reversible isothermal expansion of one mole of an ideal gas at 300 K from a volume of 10 dm³ to 20 dm³, we can follow these steps:

### Step 1: Understand the Process
In an isothermal process for an ideal gas, the temperature remains constant. For one mole of an ideal gas undergoing isothermal expansion, the change in internal energy (ΔU) is zero.

### Step 2: Use the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = \Delta H - W \]
Where:
- ΔU = change in internal energy
- ΔH = change in enthalpy
- W = work done by the system

Since ΔU = 0 for an isothermal process, we can simplify the equation to:
\[ 0 = \Delta H - W \]
Thus, we can express ΔH as:
\[ \Delta H = W \]

### Step 3: Calculate the Work Done (W)
For a reversible isothermal expansion, the work done (W) can be calculated using the formula:
\[ W = -2.303 nRT \log\left(\frac{V_2}{V_1}\right) \]
Where:
- n = number of moles (1 mole in this case)
- R = gas constant (8.314 J/mol·K)
- T = temperature in Kelvin (300 K)
- \( V_2 = 20 \, \text{dm}^3 \) (final volume)
- \( V_1 = 10 \, \text{dm}^3 \) (initial volume)

Substituting the values:
\[ W = -2.303 \times 1 \times 8.314 \times 300 \log\left(\frac{20}{10}\right) \]

### Step 4: Calculate the Logarithm
Calculate the logarithm:
\[ \log\left(\frac{20}{10}\right) = \log(2) \approx 0.301 \]

### Step 5: Substitute and Calculate W
Now substituting back into the work done equation:
\[ W = -2.303 \times 1 \times 8.314 \times 300 \times 0.301 \]
\[ W = -2.303 \times 8.314 \times 300 \times 0.301 \]
\[ W \approx -1728.8 \, \text{J} \]

### Step 6: Convert Work Done to kJ
Since we need the answer in kilojoules:
\[ W \approx -1.7288 \, \text{kJ} \]

### Step 7: Find the Enthalpy Change (ΔH)
Now substituting W back into the equation for ΔH:
\[ \Delta H = W \]
\[ \Delta H = -1.7288 \, \text{kJ} \]

Since ΔH is equal to the negative of the work done:
\[ \Delta H \approx +1.73 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change (ΔH) for the process is approximately:
\[ \Delta H \approx +1.73 \, \text{kJ} \]