Question

if 20.0g of N2 gas has a volume of 0.40L and a pressure of 6.0atm, what is the temperature in degrees celsius?

Answer 1

Answer:

-232 °C

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data:  

m = 20.0 g

p = 6.0 atm

V = 0.40 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations:

1. Moles of N₂

$\text{n} = \text{20.0 g} \times \dfrac{\text{1 mol}}{\text{28.01 g}} = \text{0.7140 mol}$

2. Temperature of N₂

$\begin{array} {rcl}pV & = & nRT\\\text{6.0 atm} \times \text{0.40 L} & = & \rm\text{0.7140 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\2.40&=&0.05859T\text{ K}^{-1}\\T& =& \dfrac{2.40 }{0.05859\text{ K}^{-1}}\\\\ & = & \text{41.0 K}\end{array}$

3. Convert the temperature to Celsius

T = (41.0 - 273.15) °C = -232 °C