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Step-by-Step Solution
Step 1: Identify the Process
The gas expands reversibly and is maintained at a constant temperature of 37°C. A reversible change at constant temperature for an ideal gas is known as an isothermal process.
Step 2: Note the Given Data
Number of moles of gas, $n = 0.04$ mol
Initial volume, $V_1 = 50.0 \text{ mL}$
Final volume, $V_2 = 375 \text{ mL}$
Temperature, $T = 37.0^\circ \text{C} = 310 \text{ K}$ (approximately)
Heat absorbed by the system, $q = +208 \text{ J}$
Gas constant, $R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}$
$\ln(7.5) = 2.01$ (given)
Step 3: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
$ \Delta U = q + w
$
For an isothermal process involving an ideal gas, the internal energy $U$ depends only on temperature. Since the temperature remains constant, $ \Delta U = 0$. Therefore:
$\Delta U = 0 \quad \Longrightarrow \quad 0 = q + w \quad \Longrightarrow \quad q = -w
$
Step 4: Express the Work Done in an Isothermal Reversible Expansion
The work $w$ done by an ideal gas in a reversible isothermal expansion from volume $V_1$ to $V_2$ is given by:
$ w = -nRT \ln\left(\frac{V_2}{V_1}\right)
$
Substituting the given values:
$
w = - (0.04 \,\text{mol}) \times (8.314 \,\text{J mol}^{-1}\text{K}^{-1}) \times (310 \,\text{K}) \times \ln\left(\frac{375}{50}\right),
$
$
\ln\left(\frac{375}{50}\right) = \ln(7.5) = 2.01,
$
$
w \approx -0.04 \times 8.314 \times 310 \times 2.01 \approx -208 \,\text{J}.
$
Step 5: Determine the Heat (q) and Work (w)
We are told the system absorbs 208 J of heat, so $q = +208 \,\text{J}$. Since from the first law $q = -w$, we have:
$
w = -208 \,\text{J}.
$
Hence,
$q = +208 \, \text{J}$
$w = -208 \, \text{J}$
Final Answer
The values of heat and work for this reversible isothermal expansion are:
q = +208 J, w = –208 J
