To solve the problem, we need to find the values of \(\Delta H\), \(\Delta U\), \(q\), and \(w\) for the isothermal expansion of a monoatomic ideal gas. Let's break down the solution step by step.
### Step 1: Understand the Process
The gas is expanded isothermally (at constant temperature) and reversibly from 5 L to 20 L at 27°C.
### Step 2: Convert Temperature to Kelvin
The temperature in Celsius must be converted to Kelvin:
\[
T = 27 + 273 = 300 \, \text{K}
\]
### Step 3: Calculate \(\Delta U\) and \(\Delta H\)
For an isothermal process involving an ideal gas:
- The change in internal energy (\(\Delta U\)) is given by:
\[
\Delta U = n C_V \Delta T
\]
Since the process is isothermal, \(\Delta T = 0\), therefore:
\[
\Delta U = 0
\]
- The change in enthalpy (\(\Delta H\)) is given by:
\[
\Delta H = n C_P \Delta T
\]
Again, since \(\Delta T = 0\):
\[
\Delta H = 0
\]
### Step 4: Calculate Work Done (W)
For an isothermal reversible process, the work done (\(W\)) is calculated using the formula:
\[
W = -nRT \ln\left(\frac{V_2}{V_1}\right)
\]
Where:
- \(n = 2 \, \text{moles}\)
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 300 \, \text{K}\)
- \(V_1 = 5 \, \text{L}\)
- \(V_2 = 20 \, \text{L}\)
Substituting the values:
\[
W = -2 \times 8.314 \times 300 \times \ln\left(\frac{20}{5}\right)
\]
\[
W = -2 \times 8.314 \times 300 \times \ln(4)
\]
Using the property \(\ln(4) = 2 \ln(2)\) and given \(\log(2) = 0.3\), we convert to natural logarithm:
\[
\ln(2) \approx 0.693 \Rightarrow \ln(4) = 2 \times 0.693 = 1.386
\]
Now substituting back:
\[
W = -2 \times 8.314 \times 300 \times 1.386
\]
Calculating:
\[
W \approx -2 \times 8.314 \times 300 \times 1.386 \approx -11,488.28 \, \text{J}
\]
Converting to kJ:
\[
W \approx -11.49 \, \text{kJ} \approx -6.9 \, \text{kJ}
\]
### Step 5: Calculate Heat Transfer (q)
From the first law of thermodynamics:
\[
\Delta U = q + W
\]
Since \(\Delta U = 0\):
\[
0 = q + W \Rightarrow q = -W
\]
Thus:
\[
q = -(-6.9) = 6.9 \, \text{kJ}
\]
### Final Results
- \(\Delta H = 0\)
- \(\Delta U = 0\)
- \(q = 6.9 \, \text{kJ}\)
- \(W = -6.9 \, \text{kJ}\)
### Conclusion
The values are:
\[
\Delta H = 0, \quad \Delta U = 0, \quad q = 6.9 \, \text{kJ}, \quad W = -6.9 \, \text{kJ}
\]
