Answer:
![\large \boxed{\text{B.) 2.8 atm}}](https://img.qammunity.org/2020/formulas/chemistry/college/mxtah8bqczuw81gcyw8i321c076gq6avcz.png)
Step-by-step explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
![(p_(1))/(T_(1)) = (p_(2))/(T_(2))](https://img.qammunity.org/2020/formulas/chemistry/middle-school/g73i995pvlhsnn4s26rdwf7gie43y1gkpa.png)
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
![\begin{array}{rcl}(1520)/(300.15) & = & (p_(2))/(423.15)\\\\5.064 & = & (p_(2))/(423.15)\\\\5.064*423.15&=&p_(2)\\p_(2) & = & \text{2143 Torr}\end{array}\\](https://img.qammunity.org/2020/formulas/chemistry/college/rpnsm8j9gby2lmxgqtzvu3ndmax6ogzn7c.png)
(c) Convert the pressure to atmospheres
![p = \text{2143 Torr} * \frac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}](https://img.qammunity.org/2020/formulas/chemistry/college/847pumq9zeqr26ebpc1ajmwgf87yexoi8c.png)