Answer :
Final answer:
The correct calculation yields a work done of 405.3 joules, which is not among the provided options.
Explanation:
The student has asked about the work done by a gas during a quasi-static isobaric expansion. In an isobaric process, the pressure (P) remains constant as the volume (V) changes. The work (W) done by the gas can be calculated using the formula W = P \\(\\Delta V), where \\(\\Delta V) is the change in volume and P is the pressure of the gas.
First, we need to convert the pressure to the correct units. We have 2.00 atm, and since 1 atm = 101.325 J/L, the pressure in these units will be P = 2.00 atm \\times 101.325 J/L = 202.65 J/L. The change in volume \\(\\Delta V) is 5.00 L - 3.00 L = 2.00 L. Now, we can find the work done:
W = P \\(\\Delta V) = 202.65 J/L \\times 2.00 L = 405.3 J
So, none of the options provided in the question (a-d) are correct, since the correct answer is 405.3 J.
