2 moles monoatomic ideal gas is expanded isothermally and reversibly from 5 L to 20 L at 27°C. `deltaH`, `deltaU`, q and w for the process respectively are (log 2 = 0.3)

Options

A. Zero, Zero, Zero and Zero
B. Zero, Zero, Zero and -6.9 kJ
C. Zero, +6.9 kJ, +6.9 kJ and -6.9 kJ
D. Zero, Zero, +6.9 kJ and -6.9 kJ

Answer

To solve the problem step by step, we will calculate the change in enthalpy (ΔH), change in internal energy (ΔU), heat (q), and work done (w) for the isothermal expansion of a monoatomic ideal gas.

### Step 1: Understand the process
We have 2 moles of a monoatomic ideal gas expanding isothermally and reversibly from 5 L to 20 L at a temperature of 27°C (which is 300 K).

### Step 2: Calculate ΔU
For an isothermal process involving an ideal gas, the change in internal energy (ΔU) is given by:
\[ \Delta U = nC_V\Delta T \]
Since the process is isothermal, ΔT = 0. Therefore:
\[ \Delta U = 0 \]

### Step 3: Calculate ΔH
For an ideal gas, the change in enthalpy (ΔH) is given by:
\[ \Delta H = nC_P\Delta T \]
Again, since the process is isothermal (ΔT = 0):
\[ \Delta H = 0 \]

### Step 4: Calculate work done (w)
The work done during an isothermal expansion of an ideal gas can be calculated using the formula:
\[ w = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
Where:
- \( n = 2 \) moles
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \)
- \( V_f = 20 \, \text{L} \)
- \( V_i = 5 \, \text{L} \)

Substituting the values:
\[ w = -2 \times 8.314 \times 300 \times \ln\left(\frac{20}{5}\right) \]
\[ = -2 \times 8.314 \times 300 \times \ln(4) \]
Using the approximation \(\ln(4) = 2 \ln(2) = 2 \times 0.3 = 0.6\):
\[ w = -2 \times 8.314 \times 300 \times 0.6 \]
Calculating this gives:
\[ w = -2 \times 8.314 \times 300 \times 0.6 = -2997.84 \, \text{J} \approx -6.83 \, \text{kJ} \]

### Step 5: Calculate q
From the first law of thermodynamics:
\[ \Delta U = q + w \]
Since ΔU = 0:
\[ 0 = q + w \implies q = -w \]
Thus:
\[ q = 6.83 \, \text{kJ} \]

### Summary of Results
- ΔU = 0
- ΔH = 0
- w = -6.83 kJ
- q = 6.83 kJ

### Final Answer
The values for the process are:
- ΔH = 0
- ΔU = 0
- q = 6.83 kJ
- w = -6.83 kJ