If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands from 6.00 L to 18.00 L and has a final pressure of 1.20 atm, isothermally and reversibly, Which of the following is correct?

Question

Quizplus · Accepted Answer

For an isothermal, reversible expansion of an ideal gas, the work done (w) is given by $w = -nRT \ln\left(\frac{V_f}{V_i}\right)$, where $n$ is the number of moles, $R$ is the gas constant (8.314 J/mol·K), $T$ is the temperature in Kelvin, $V_f$ is the final volume, and $V_i$ is the initial volume. Since the process is isothermal, $\Delta U = 0$ (change in internal energy for an ideal gas is zero), and the heat absorbed (q) is equal in magnitude but opposite in sign to the work done on the system, making $q = -w$. Calculating the work and recognizing that $\Delta U = 0$ for an isothermal process in an ideal gas confirms choice A as correct.

If 200 Mol of an Ideal Gas at 300 K and and 3.00