Question

if 3.07 mil of an ideal gas has a pressure of 2.91 atm and a volume of 78.13 liters, what is the temperature of the sample in celsius

Answer 1

Answer:

$\large \boxed{\text{629 ^{\circ} C}}$

Explanation:

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

pV = nRT

Data  

p = 2.91 atm

V = 78.13 L

n = 3.07 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

$\begin{array} {rcl}pV & = & nRT\\\text{2.91 atm} \times \text{78.13 L} & = & \rm\text{3.07 mol} \times 0.08206 \text{L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\227.4&=&0.2519T\text{ K}^{-1}\\T& = &\dfrac{227.4}{\text{0.2519 K}^{-1}}\\\\ & = & \text{902.5 K}\\\end{array}\\T = \text{(902.5 - 273.15) ^{\circ} C} = \large \boxed{\textbf{629 ^{\circ} C}}$