Question

A piston filled with $0.04mol$ of an ideal gas expands reversibly from $50mL$ to $375mL$ at a constant temperature of $37.0^\circ C$. As it does so, it absorbs $208\,J$ of heat. The values of $q$ and $w$ for the process will be : $(R = 8.314J/molK)$

Answer 1

A reversible isothermal expansion is an infinitely slow increase in volume $V$ at constant pressure and temperature. All the reversible isothermal $PV$ work ${W_{rev}}$ done by an ideal gas to expand was possible by reversibly absorbing heat ${q_{rev}}$ into the ideal gas.

Complete step-by-step solution: For isothermal reversible expansion of an ideal gas, you can write the following expression to calculate the work done
$W = 2.303 \times n \times R \times T \times \log \left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)$
Substitute 0.04 for n (the number of moles), 8.314 for R(the ideal gas constant), 310 for T (absolute temperature in Kelvin), 50 and 375 for V1 and V2 (the initial and final volume) in the above expression and calculate the work done.
$W = 2.303 \times 0.04 \times 8.314 \times \left( {37 + 273.15} \right) \times \log \left( {\dfrac{{50}}{{375}}} \right)$
$W = - 208\,J$
Hence, the work done in the reversible isothermal process is $- 208\,J$ .

The $1st$ law of thermodynamics states the conservation of energy principle to systems where heat transfer and work done are the methods of transferring energy inside the system and outside of the system. The $1st$ law of thermodynamics also states that the difference in the internal energy of a system equals the subtraction net amount of heat transfer inside the system and the net amount of the work done by the system. To write it in an equation form, the $1st$ law of thermodynamics is denoted as, $\Delta U = Q - W$.
For isothermal process,$\Delta U = 0$
Change of state or phase changing of different liquids through the total process of melting and evaporation of that liquid are examples of the isothermal process. One of the examples of the industrial application of the isothermal process is the Carnot engine and refrigerator works isothermally.
As mentioned in the question the amount heat is absorbed,
$q = + 208\,J$
Now for reversible Isothermal reaction process,
$q = - W$
$W = - q = - 208\,J$.
Hence the values of q and W for the process will be $+ 208\,J$ and $- 208\,J$ respectively.

Therefore, the correct answer is $q = 208\,J,W = - 208\,J$.

Note: Reversible work done determines that the entire system (including all the surrounding system) can be get back to the starting state and the irreversible work done is the process that we can only bring the piston return to the starting state if the surrounding system is changed.