A piston filled with 0.04 mole of an ideal gas expands reversible from 50.0 mL to 375 mL at a constant temperature of `37.0^(@)C`. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: (R=3.14J/molK) (in 7.5=2.01)

Options

A. q=+208J, w=-208 J
B. q=-208J,w=-208J
C. q=-208 J, w=++209J
D. q=+208J, w=+208J

Explanation

To solve the problem, we need to find the values of \( q \) (heat absorbed) and \( w \) (work done) for the isothermal reversible expansion of an ideal gas.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Number of moles of gas, \( n = 0.04 \, \text{moles} \)
- Initial volume, \( V_1 = 50.0 \, \text{mL} = 0.050 \, \text{L} \)
- Final volume, \( V_2 = 375.0 \, \text{mL} = 0.375 \, \text{L} \)
- Temperature, \( T = 37.0^\circ C = 310.15 \, \text{K} \) (convert to Kelvin)
- Heat absorbed, \( q = 208 \, \text{J} \)
- Ideal gas constant, \( R = 3.14 \, \text{J/mol K} \)

2. **Determine Change in Internal Energy**:
- For an isothermal process involving an ideal gas, the change in internal energy \( \Delta U = 0 \).

3. **Apply the First Law of Thermodynamics**:
- The first law states:
\[ \Delta U = q + w \]
- Since \( \Delta U = 0 \), we can rearrange this to find \( w \):
\[ 0 = q + w \implies w = -q \]

4. **Calculate Work Done (w)**:
- Substitute the value of \( q \):
\[ w = -208 \, \text{J} \]

5. **Final Values**:
- The values are:
- \( q = 208 \, \text{J} \)
- \( w = -208 \, \text{J} \)

### Summary of Results:
- \( q = 208 \, \text{J} \)
- \( w = -208 \, \text{J} \)