A gas undergoes two processes. In the first, the volume remains constant at 0.200 m^3 and the pressure increases from 2.00 * 10^5 Pa to 5.00 * 10^5 Pa. The second process is a compression to a volume of 0.120 m^3 at a constant pressure of 5.00 * 10^5 Pa. (b) Find the total work done by the gas during both processes.
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1
Identify the type of processes involved. The first process is isochoric (constant volume), and the second process is isobaric (constant pressure).
Recall the formula for work done by a gas, which is W = P \\Delta V, where P is the pressure and \\Delta V is the change in volume. For an isochoric process, since the volume does not change, \\Delta V = 0, and thus no work is done (W = 0).
Calculate the change in volume for the second process. Since it is an isobaric process, use the initial and final volumes to find \\Delta V = V_{final} - V_{initial} = 0.120 m^3 - 0.200 m^3.
Substitute the values into the work formula for the second process. Use the constant pressure and the change in volume calculated in the previous step.
Add the work done in both processes to find the total work done by the gas. Since the work done in the first process is zero, the total work done is just the work done during the second process.