Consider a piston filled with 1 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting at 2 m3 and ending at 1 m3 How much work do we need to do on the gas to perform this operation? A. -1670 J B. 1670 JC. -2410 J D. 2410 J E. No work is done because the temperature is constant.

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W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j