Answer:
[tex]\large \boxed{\text{B) 84 mL}}[/tex]
Explanation:
We can use the Combined Gas Laws to solve this problem
.
[tex]\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}[/tex]
Data
p₁ = 0.75 atm; V₁ = 120.0 mL; T₁ = 295 K
p₂ = 1.25 atm; V₂ = ?; T₂ = 345 K
Calculations
[tex]\begin{array}{rcl}\dfrac{p_{1}V_{1} }{T_{1}} & = & \dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{\text{0.75 atm $\times$ 120.0 mL}}{\text{295 K}} & = & \dfrac{\text{1.25 atm} \times V_{2}}{\text{345 K}}\\\\\text{0.305 mL} & = & \text{0.003 523V}_{2}\\V_{2}& =& \dfrac{\text{0.305 mL}}{0.003523}\\\\& = & \textbf{84 mL}\\\end{array}\\\text{The new volume of the gas is $\large \boxed{\textbf{84 mL}}$}[/tex]
