# If 20.0 g of N_2 gas has a volume of 0.40 L and a pressure of 6.0 atm, what is its Kelvin temperature?

Jun 28, 2018

41 L

#### Explanation:

This is a gas law question, and to decide which gas law to use, look at the information that is given. It looks like there is only one set of data given, meaning only 1 volume, 1 pressure values. That means, we'll be using the Ideal Gas Law to solve this problem. The formula is,

$p V = n R T$ where

$R = \text{gas constant" =0.0821"L*atm"/"K*mol}$

There are various values for the gas constant, but we'll use this one due to the unit of the pressure and volume that is given.

From the question, we have
$p = 6.0 \text{atm}$
$V = 0.40 \text{L}$
$m a s s = 20.0 \text{g}$
and we need to find $T$

From the formula, we lack n, which is number of moles of ${N}_{2}$. We can find that using the mass of ${N}_{2}$ that is given.

$\text{moles of "N_2, n= "mass"/"molar mass"="20.0g"/"(2x14)g/mol"=0.7143 "mols}$

$p V = n R T$
$\left(6.0 \text{atm")(0.40L)=(0.7143mols)(0.0821"L*atm"/"K*mol}\right) \cdot T$

$T = \text{(6.0atm)(0.40L)"/"(0.7143mols)(0.0821L*atm/K*mol)} = 40.92 K$

Final answer is 41 L (2 significant figures)