Question

A 3.0 mol gas sample occupies 6.0 liters at 25 °C. What is the pressure of the gas in kPa’s?

Answer 1

The pressure will be quite high!

Explanation:

`P=(nRT)/V`

`=` `{(3.0*cancel(mol)*0.0821*cancel(L)*atm*cancel(K^(-1))cancel(mol)^(-1)xx298cancelK)/(6.0*cancelL)}xx101.3*kPa*(atm)^-1`

Is the pressure you get sensible? Would you a priori expect it to be greater or lesser than atmospheric pressure?

Answer 2

The pressure will be `"1240 kPa"`

Explanation:

Use the ideal gas law with the equation `PV=nRT`, where `P` is pressure, `V` is volume, `n` is moles, `R` is the gas constant, and `T` is the Kelvin temperature.

Given/Known
`V="6.0 L"`
`n="3.0 mol"`
`R="8.3144598 L kPa K"^(-1) "mol"^(-1)"`
https://en.wikipedia.org/wiki/Gas_constant
`T="25"^"o""C"+273.15="298 K"`

Unknown
pressure, `P`

Equation
`PV=nRT`

Solution
Rearrange the equation to isolate `P` and solve.

`P=(nRT)/V`

`P=(3.0cancel"mol"xx8.3144598cancel"L" "kPa" cancel("K"^(-1)) cancel("mol"^(-1))xx298cancel"K")/(6.0cancel"L")="1240 kPa"` (rounded to three significant figures) http://academic.umf.maine.edu/magri/PUBLIC.acd/tools/SigFigsAndRounding.html