Question

If 2.19 mol

Answer 1

Answer:

1370 °C

Explanation:

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

pV = nRT

Data:  

p = 8.12 atm

V = 36.41 L

n = 2.19

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

$\begin{array} {rcl}pV & = & nRT\\\text{8.12 atm} \times \text{36.41 L} & = & \rm\text{2.19 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\295.6&=&0.1797T\text{ K}^{-1}\\T& = &\dfrac{294.9 }{\text{0.1797 K}^{-1}}\\\\ & = & \text{1645 K}\\\end{array}$

T = (1645 - 273.15) °C = 1370 °C

The temperature of the sample is 1370 °C.