A 3.6 mole sample of methane gas is kept in a 1.50 liter container at a temperature of 100°C. What is the pressure of the gas?

Mar 25, 2018

I get $7442.7 \setminus \text{kPa}$.

Explanation:

We use the ideal gas law, which states that,

$P V = n R T$

• $P$ is the pressure

• $V$ is the volume in liters (for this case)

• $n$ is the number of moles of the substance

• $R$ is the ideal gas constant, which varies

• $T$ is the temperature in Kelvin

Since we need to find pressure, we can rearrange the equation into:

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

Now, we need to convert the temperature into Kelvin. We know that $\text{K"=""^@"C} + 273.15$, and so ${100}^{\circ} \text{C"=100+273.15=373 \ "K}$.

Since our temperature is in $\text{K}$ and volume in liters, let's use $R = 8.314 \setminus {\text{L" \ "kPa" \ "K"^-1 \ "mol}}^{-} 1$. Taken from: https://en.wikipedia.org/wiki/Gas_constant

And so, we find that the pressure is:

$P = \left(3.6 \setminus \text{mol"*8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1*373 \ "K")/(1.5 \ "L}\right)$

$= \left(11164.0392 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"color(red)cancelcolor(black)"L" \ "kPa"color(red)cancelcolor(black)"K"^-1color(red)cancelcolor(black)"mol"^-1color(red)cancelcolor(black)"K")/(1.5color(red)cancelcolor(black)"L}}}}\right)$

$= 7442.6928 \setminus \text{kPa}$

$\approx 7442.7 \setminus \text{kPa}$