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

Dec 2, 2015

The pressure will be quite high!

Explanation:

$P = \frac{n R T}{V}$

$=$ $\left\{\frac{3.0 \cdot \cancel{m o l} \cdot 0.0821 \cdot \cancel{L} \cdot a t m \cdot \cancel{{K}^{- 1}} {\cancel{m o l}}^{- 1} \times 298 \cancel{K}}{6.0 \cdot \cancel{L}}\right\} \times 101.3 \cdot k P a \cdot {\left(a t m\right)}^{-} 1$

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

Dec 2, 2015

The pressure will be $\text{1240 kPa}$

Explanation:

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

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

Unknown
pressure, $P$

Equation
$P V = n R T$

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

$P = \frac{n R T}{V}$

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