Answer:
[tex]\large \boxed{\text{-92 $^{\circ}$C}}[/tex]
Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data
p = 1.25 atm
V = 25.0 L
n = 2.10 mol
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations
1. Temperature in kelvins
[tex]\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}[/tex]
2. Temperature in degrees Celsius
[tex]\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}[/tex]
