To solve the problem, we need to determine the values of \( q \) (heat absorbed) and \( w \) (work done) for the isothermal expansion of the ideal gas.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Number of moles, \( n = 0.04 \) mol
- Initial volume, \( V_1 = 50.0 \) mL = \( 0.050 \) L
- Final volume, \( V_2 = 375 \) mL = \( 0.375 \) L
- Temperature, \( T = 37.0^\circ C = 310.15 \) K (convert to Kelvin by adding 273.15)
- Heat absorbed, \( q = +208 \) J
2. **Determine the Work Done (w):**
- For an isothermal process involving an ideal gas, the work done by the gas during expansion can be calculated using the formula:
\[
w = -nRT \ln\left(\frac{V_2}{V_1}\right)
\]
- First, calculate \( \frac{V_2}{V_1} \):
\[
\frac{V_2}{V_1} = \frac{0.375 \, \text{L}}{0.050 \, \text{L}} = 7.5
\]
- Now, substitute the values into the work formula:
\[
w = - (0.04 \, \text{mol}) \times (8.314 \, \text{J/mol K}) \times (310.15 \, \text{K}) \times \ln(7.5)
\]
- Given \( \ln(7.5) \approx 2.01 \):
\[
w = - (0.04) \times (8.314) \times (310.15) \times (2.01)
\]
- Calculate \( w \):
\[
w \approx - (0.04) \times (8.314) \times (310.15) \times (2.01) \approx - 208 \, \text{J}
\]
3. **Determine \( q \):**
- Since the process is isothermal, we know from the first law of thermodynamics that:
\[
\Delta U = q + w
\]
- For an isothermal process, the change in internal energy \( \Delta U = 0 \):
\[
0 = q + w
\]
- Rearranging gives:
\[
q = -w
\]
- Substituting the value of \( w \):
\[
q = -(-208 \, \text{J}) = 208 \, \text{J}
\]
4. **Final Values:**
- \( q = +208 \, \text{J} \)
- \( w = -208 \, \text{J} \)
### Summary:
- The heat absorbed \( q \) is \( 208 \, \text{J} \).
- The work done \( w \) is \( -208 \, \text{J} \).
