An ideal gas system is maintained at a constant volume of 4 l. if the pressure is constant, how much work is done by the system?

For #1: If the volume is unchanged no work is done since nothing was moved outside the system or inside the system. The equation is W = P (delta)V, or work equals pressure times the change in volume. If the change in volume is zero, so is the work. 

For #2: If 200 joules is added then this must be the change in internal energy. Since there was no other addition and there is no mention of work being done on or by the gas, the 200 J of heat added 200 J to the internal energy of the system. No equation necessary for this situation. 

For #3: No work is done ON the system because the system is doing work on the environment. In other words, the volume is expanding so the system is pushing outward. The situation says that the system is in a vacuum, so there is nothing for the system to push on except itself, that is, the container that did the expanding. Since this is not a gas, it would not be considered as part of the system. 

For #4: I do not know how the answer 0.33 was arrived at. The equation I use to calculate the efficiency of an engine is e = 1 – Q(low) / Q(high). As you can see, this gives an efficiency of 50%. H . . . . 

I hope this was at least a little help.

The charge of an ion with 55 protons and 52 electrons is +3. the reason is because protons are positive, and there are three more than the amount of negative electrons.