Question

Question #d4c35

Answer 1

`"0.700 g He"`

Explanation:

The idea here is that when the pressure and the temperature of a gas are kept constant, the volume of the gas and the number of moles present in the sample have a direct relationship described by Avogadro's Law.

In other words, under these conditions, when you increase the number of moles of gas present in the sample, the volume will increase as well. Similarly, when you decrease the number of moles of gas present in the sample, the volume will decrease as well.

Mathematically, you can write this as

`V_1/n_1 = V_2/n_2`

Here

• `V_1`, `n_1` represent the volume and the number of moles of gas at an initial state
• `V_2`, `n_2` represent the volume and the number of moles of gas at a final state

Now, the first thing that you need to do here is to figure out how many moles of helium are present in the cylinder when the volume of the gas is equal to `"2.70 L"`.

Rearrange the equation to solve for `n_2`

`n_2 = V_2/V_1 * n_1`

As you know, helium has a molar mass of `"4.0026 g mol"^(-1)`, which means that the initial number of moles present in the cylinder can be calculated by dividing the mass of the sample by the molar mass of helium.

`n_1 = (2.00 color(red)(cancel(color(black)("g"))))/(4.0026color(red)(cancel(color(black)("g"))) "mol"^(-1)) = "0.4997 moles"`

You can thus say that you have

`n_2 = (2.70 color(red)(cancel(color(black)("L"))))/(2.00color(red)(cancel(color(black)("L")))) * "0.4997 moles"`

`n_2 = "0.6746 moles"`

This means that the number of moles of helium increased by

`Deltan = n_2 - n_1`

`Deltan = "0.6746 moles " - " 0.4997 moles"`

`Deltan = "0.1749 moles"`

To convert this to grams, use the molar mass of helium

`0.1749 color(red)(cancel(color(black)("moles He"))) * "4.0026 g"/(1color(red)(cancel(color(black)("mole He")))) = color(darkgreen)(ul(color(black)("0.700 g")))`

I'll leave the answer rounded to three sig figs.