A piston filled with 0.04 mol of an ideal gas expands reversibly 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 = 8.314 J/mol K) (ln7.5 = 2.01)

We have a reversible, isothermal expansion of an ideal gas. For an ideal gas, the internal energy $$\Delta U$$ depends only on temperature, and because the temperature is held constant, we can state immediately

$$\Delta U = 0.$$

The First Law of Thermodynamics gives the relation

$$\Delta U = q + w,$$

where $$q$$ is the heat absorbed by the system and $$w$$ is the work done by the system (work is taken as negative when the system does work on the surroundings in the convention normally used in chemistry).

Since $$\Delta U = 0$$, the First Law simplifies to

$$0 = q + w \quad\Rightarrow\quad q = -w.$$

Thus, once we calculate the reversible isothermal work, we can immediately find $$q$$ from the above relation.

For a reversible, isothermal expansion of an ideal gas the expression for work is

$$w = -nRT \ln\!\left(\dfrac{V_f}{V_i}\right),$$

where

$$n = 0.04\$$ mol $$, \quad R = 8.314\$$ J mol $$^{-1}$$ K $$^{-1}, \quad T = 37.0^\circ$$ C $$= 37 + 273 = 310\$$ K $$,$$

and

$$V_i = 50.0\$$ mL $$, \qquad V_f = 375\$$ mL $$.$$

Now we form the volume ratio:

$$\dfrac{V_f}{V_i} = \dfrac{375\$$ mL $$}{50.0\$$ mL $$} = 7.5.$$

The problem statement supplies $$\ln 7.5 = 2.01.$$ Substituting everything into the work formula, we obtain

$$w = -\bigl(0.04\$$ mol $$\bigr)\bigl(8.314\$$ J mol $$^{-1}$$ K $$^{-1}\bigr)\bigl(310\$$ K $$\bigr)\bigl(\ln 7.5\bigr).$$

First we multiply the gas constant by the temperature:

$$8.314 \times 310 = 2577.34\ \text{J mol}^{-1}.$$

Next we multiply by the number of moles:

$$0.04 \times 2577.34 = 103.09\ \text{J}.$$

Finally we incorporate the logarithm:

$$w = -103.09 \times 2.01 \approx -207.2\ \text{J}.$$

The value is essentially $$-208\ \text{J}$$ when rounded to three significant figures, which matches the precision of the heat value given in the problem.

With the work known, we return to the simplified First Law relation $$q = -w$$ to find the heat absorbed:

$$q = -(-208\ \text{J}) = +208\ \text{J}.$$

The positive sign shows that the system absorbs 208 J of heat, exactly as stated in the problem text. Hence our calculated signs are self-consistent:

$$q = +208\ \text{J}, \qquad w = -208\ \text{J}.$$

Therefore the correct option is the one listing $$q = +208\ \text{J}$$ and $$w = -208\ \text{J}.$$

Hence, the correct answer is Option C.