Calculate the volume occupied by 56.5 g of argon gas at stp

Molar mass of argon = 39.95 grams
number of moles of Argon = mass / molar mass 
number of moles of Argon = 56.5 / 39.95 = 1.414 moles

At STP, one mole of gas occupies 22.4 liters. To know the volume that 1.414 moles occupy, all you have to do is cross multiplication as follows:
volume = (1.414*22.4) / 1 = 31.679 liters

31.7 liters if using the pre 1982 definition of STP. 32.1 liters if using the post 1981 definition of STP. First, determine how many moles of argon you have by dividing the mass you have by the atomic weight of argon. So 56.5 g / 39.948 g/mol = 1.41433864 mol Second, determine what definition of STP you're using. the definition prior to 1982 of STP was 0°C with a pressure of 1 atmosphere (1.01325 X 10^5 Pascals). The definition for 1982 and later is 0°C with a pressure of 1 X 10^5 Pascals. There are still quite a few textbooks that use the older definition of STP so I'll be giving the results for both the old and new definitions of STP. To calculate the volume of an ideal gas, just multiply the number of moles by a constant. That constant is 22.710980 liters/mol for the 1982 and later standard and 22.414 liters/mol for the pre 1982 standard. So Pre 1982 1.414339 * 22.414 = 31.70098628 liters = 31.7 liters post 1981 1.414339 * 22.71098 = 32.12101657 liters = 32.1 liters

Work done = force ×distancebut force = mass ×gravityand gravity = 10m\s force = 2×10 = 20 work done = 20×3 = 60 joules

I’m not sure what your trying to say here.. can you clarify again?