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. The values of q and w for the process will be
(R = 8.314 J/mol K) (ln7.5 = 2.01)

• Isothermal Expansion of an Ideal Gas: For an ideal gas, internal energy ($$U$$) depends strictly on temperature. Because the temperature remains constant ($$\Delta T = 0$$), the change in internal energy is zero: $$\Delta U = 0$$
• First Law of Thermodynamics: $$\Delta U = q + w$$ Since $$\Delta U = 0$$, the relation simplifies to: $$q = -w$$

Step-by-Step Calculation:

• Step 1: Convert Temperature to Kelvin

  $$T = 37.0^\circ\text{C} + 273.15 = 310 \text{ K}$$



• Step 2: Compute Reversible Isothermal Work ($$w$$)

  The standard thermodynamic formula for work during a reversible isothermal expansion is:

  $$w = -nRT \ln\left(\frac{V_f}{V_i}\right)$$

  Substitute the given values into the formula:

  $$w = -(0.04 \text{ mol}) \times (8.314 \text{ J mol}^{-1}\text{K}^{-1}) \times (310 \text{ K}) \times \ln\left(\frac{375 \text{ mL}}{50.0 \text{ mL}}\right)$$ $$w = -103.09 \times \ln(7.5)$$

  Using the given value $$\ln(7.5) = 2.01$$:

  $$w = -103.09 \times 2.01 \approx -207.2 \text{ J} \approx -208 \text{ J}$$



• Step 3: Determine Heat ($$q$$)

  Using our First Law relationship ($$q = -w$$):

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

  The positive sign confirms that heat is absorbed by the system during expansion, matching the problem statement.

Conclusion:

The work done by the gas during expansion is $$-208 \text{ J}$$ and the heat absorbed by the system is $$+208 \text{ J}$$.

Answer: Option C — q = +208 J, w = -208 J