Question in the description can you please explain how you got the answer, thank you.

Question

A cylinder with a movable piston contains 6.00 g of helium, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 5.00 L to 6.50 L? (The temperature was held constant.)

Answer 1

Since the pressure and temperature are held constant, then the only thing affecting the change in volume is the addition of the extra helium gas. A total of 1.50 Liters (6.50 L - 5.00 L) of helium is added.

Avogadro's Law tells us that the volume of a gas is directly proportional to the number of moles of a gas (22.414 L of any gas = 1 mole of gas) and we can also use the molar mass of helium (4.00 g/mol, from the Periodic Table of the Elements) to turn the moles of gas into a mass.

Setting up the problem the goal is to set up ratios that allow units to cancel out.

1.5 L He × 1 mol He/22.414 L He × 4.00 g He / 1 mole He = 0.267689837 g He ≈ 0.27 g He

Lastly, we round the answer to have the same number of significant figures [digits] as the numbers given in the problem (all the values go to the hundredths place, so our answer should be rounded to that place also).

The only additional complicating factor is that Avogadro's Law applies to gases at STP [standard temperature (0°C or 273.15 K) and pressure (1 atm or 101.3 kPa [there are other units possible too])]. However, since the specifics of room temperature aren't given in this problem, it is probably negligible, but I will work it out using 20°C [273.15 K + 20°C = 293.15 K, converts the temperature to Kelvin since we have to use an absolute [no negative numbers] scale for chemistry calculations] as room temperature to show that conversion too.

Using the common gas law, P_bV_b/T_b = P_aV_a/T_a (where _b is _{before or standard conditions} and _a is _{after or lab/problem conditions}, and since the Pressure is held constant it can be ignored [V_b/T_b = V_a/T_a] then:

1.5 L / 273.15 K = V_a/293.15 K , rearranging algebraically by multiplying both sides by 293.15 K to cancel out the denominator on the right leaving V_a alone we get: 1.5 L × 293.15 K / 273.15 K = V_a , doing the math (and cancelling out the K units) we get 1.60982976 L He, so we would plug this value into the above calculation in place of the starting 1.50 L.

1.60982976 L He × 1 mol He/22.414 L He × 4.00 g He / 1 mole He = 0.0179556277 g He ≈ 0.018 g He

There are other ways to solve this problem if the values of pressure and/or [room] temperature had been given. It would have also been possible to combine the PV/T=PV/T calculation with the unit conversion calculation, but I've found that it makes more sense to most students to take it in two steps, rather than trying to memorize other conversion factors.