Well, before we discuss that, I think we have to carefully understand and agree on something. We have to be very clear about what we mean by 'weight' ... is it what you feel, or is it the product of
(your mass) x (the acceleration of gravity where you are).
If you're on a space ship, then any time your engine is not burning, you feel weightless. It doesn't matter where you are, or what body you may be near. If you're not doing a burn, and the only force on you is the force of gravity, then you don't feel any weight at all.
But of we say that your 'weight' is the product of
(your mass) times (the acceleration of gravity where you are),
then it depends on where you are, and whether you're close to the Earth or closer to the moon. You may not feel it, but you're going to have weight, and it's going to change during your trip in space.
You know that the force of gravity depends on how far you are from the body that's attracting you.
-- As you travel from the Earth to the moon, gravity will pull you less and less toward Earth, and more and more toward the moon.
-- Your weight will get less and less, until you reach the point in space where the gravitational attractions are equal in both directions. That's about 24,000 miles before you reach the moon ... about 90% of the way there. At that point, your weight is really zero, because the pull toward the Earth and the pull toward the moon are equal.
-- From there, the rest of the way to the moon, your weight will start to grow again. It begins at zero at the 'magic point', and it grows and grows until you reach the moon's surface. When you're there, your weight has grown to about 1/6 of what you weigh on Earth, and it won't get any bigger. If you weigh 120 pounds on Earth, then you weigh about 19.86 pounds on the moon ... PLUS your space suit, boots, heater/air conditioner, oxygen tank, radiation shielding, radio, and all the other stuff that you need to survive on the moon for a few hours.
At the entrance of most beaches, there is a bulletin board with notices about water conditions: maybe a faded sign warning about rip currents and a list of this week's tide tables. Most people pass them by without a second thought, but if you want to enter the ocean, it is important to know its movements, whether to avoid being caught in a riptide or to figure out when the waves will be at their best.
You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J
