Question

When the temperature of a rigid hollow sphere containing 685 L of He gas is held at 348OC, the pressure of the gas is 1.89 x 103 kPa. How many moles of He does the sphere contain?

Answer 1

`"251 mol"`.

Explanation:

(Assuming you meant `1.89 xx 10^3` kPa as the pressure.)

For this question, we're given:

• `P`, or pressure, which is `1.89 xx 10^3 kPa`. This is the same as `(1.89 xx 10^3) / 101.325 = "18.65 atm"`.
• `V`, or volume, which is `"685 L"`.
• `T`, or temperature, which is `348 °C`. This is the same as `348 + 273.15 = "621.15 K"`.

Assuming that helium behaves as an ideal gas, we can use the Ideal Gas Law:

`pV=nRT`

We can also rearrange this to have only `n`, or number of moles on one side:

`n = (pV)/(RT)`

Let's say that the value of `R`, the universal gas constant, is `"0.08206 L atm / K mol"` in this case.
Plugging in all of the values, we get:

`n = (pV)/(RT)`
`n = ("18.65 atm" xx "685 L")/("0.08206 L atm / K mol" xx "621.15 K")`
`n = ("18.65" cancel("atm") xx "685" cancel("L"))/("0.08206" cancel("L atm") "/" cancel("K") mol xx 621.15 cancel("K"))`
`n = "251 mol"`