Question

What volume (L) will 0.20 mol HI occupy at 300 K and 100.0 kPa? R = 8.314 kPa L / (K mol) = 0.08205 atm L / (mol K)?

Answer 1

The answer is `5L`.

You can use the ideal gas law, `PV = nRT`, to determine the volume, given the fact that all other variables are known.

First, let's use the equation that has `R = 0.08205 (atm * L)/(mol * K)`. This means that we need to convert the `100.0kPa` pressure to `atm`

`100.0 kPa * (1 atm)/(101.325 kPa) = 0.9869atm`

So, `V = (nRT)/P = (0.2 * 0.08205 * 300)/(0.9869) = 5L` (rounded to 1 sig fig);

As practice and as a way to double-check our result, use the other value of `R` as well; this time, the pressure stays in `kPa` and the volume must come out to be `5L`

`V = (nRT)/P = (0.2 * 8.314 * 300)/(100.0) = 5L`