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

-232 °C

Step-by-step 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} * \frac{\text{1 mol}}{\text{28.01 g}} = \text{0.7140 mol}`

2. Temperature of N₂

`\begin{array} {rcl}pV & = & nRT\\\text{6.0 atm} * \text{0.40 L} & = & \rm\text{0.7140 mol} * 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1) * T\\2.40&=&0.05859T\text{ K}^(-1)\\T& =& \frac{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