Answer:
[tex]\large \boxed{\text{490 $^{\circ}$C}}[/tex]
Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data
p = 28 atm
V = 0.045 L
n = 0.020 mol
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations
[tex]\begin{array} {rcl}pV & = & nRT\\\text{28 atm} \times \text{0.045 L} & = & \rm\text{0.020 mol} \times 0.08206 \text{L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\1.26&=&0.001 641T\text{ K}^{-1}\\T& = &\dfrac{1.26}{\text{0.001 641 K}^{-1}}\\\\\end{array}[/tex]
[tex]\begin{array} {rcl} & = & \text{767.7 K}\\\end{array}\\T = \text{(767.7 - 273.15) $^{\circ}$C} = \large \boxed{\textbf{490 $^{\circ}$C}}[/tex]
