# How many moles of gas are contained in .89 L at 21.0 oC and .987 atm pressure?

## How many moles of gas are contained in .89 L at 21.0 oC and .987 atm pressure?

Mar 3, 2018

$0.03609 m o l$

#### Explanation:

We shall assume that this is an Ideal Gas, and can thus use the Ideal Gas Law:

$P V = n R T$

where $P =$ Pressure (pascals)

where $V =$ Volume (${m}^{3}$)

where $n =$ moles (mol)

where $T =$ Temperature (Kelvin)

and $R$ is the gas constant $8.31441 J {K}^{-} 1 m o {l}^{-} 1$

Now, we convert the units:

$0.987 a t m = 100007.775 p a$

$0.89 L = 0.89 {\mathrm{dm}}^{3} = 0.00089 {m}^{3}$

${21}^{o} C = 294.15 K$

Inputting the values into the equation:

$\to 100007.775 \cdot 0.00089 = n \cdot 8.31441 \cdot 294.15$

$\to 89.0069 = 2445.6837 n$

Solve for n:

$n = \frac{89.0069}{2445.6837}$

$n = 0.03639 m o l$

Thus, solved.