Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

In the first pathway, the gas undergoes an isothermal expansion from state A to state B. Therefore, we can use the formula mentioned above: $$ W_1 = -P_{A} V_{A} \ln \frac{V_{B}}{V_{A}} $$ Plugging in the given values into the formula: $$ W_1 = -\left(2.00\,\text{atm}\right)\left(10.0\,\text{L}\right) \ln \frac{30.0\,\text{L}}{10.0\,\text{L}} $$ To convert the work from L.atm to J, we can use the conversion factor: 1 L.atm ≈ 101.3 J. Therefore, we have: $$ W_1 = -\left(2.00\,\text{atm}\right)\left(10.0\,\text{L}\right) \ln 3 \times 101.3\,\frac{\text{J}}{\text{L.atm}} $$ After calculation, we get: $$ W_1 \approx -676.92\,\text{J} $$

In the second pathway, the gas first undergoes an isothermal compression at constant volume, followed by an isothermal expansion at constant pressure. Since the volume does not change during the first part of the process, no work is done. Therefore, we only need to calculate the work done during the second part, where the gas expands at constant pressure: $$ W_2 = -P_{B} V_{B} \ln \frac{V_{B}}{V_{A}} $$ Plugging in the given values into the formula: $$ W_2 = -\left(1.00\,\text{atm}\right)\left(30.0\,\text{L}\right) \ln \frac{30.0\,\text{L}}{10.\,\text{L}} $$ Multiplying by the conversion factor from L.atm to J: $$ W_2 = -\left(1.00\,\text{atm}\right)\left(30.0\,\text{L}\right) \ln 3 \times 101.3\,\frac{\text{J}}{\text{L.atm}} $$ After calculation, we get: $$ W_2 \approx -338.46\,\text{J} $$

A state function is a property that depends only on the initial and final states of a system, and not on the path taken to reach those states. Since the work done is different for the two pathways, even though the initial and final states are the same, work is not a state function.