Answer:
[tex]58.48^{\circ}\text{C}[/tex]
Explanation:
[tex]P_1[/tex] = Initial pressure = 2.77 atm
[tex]V_1[/tex] = Initial volume = 263 mL
[tex]T_1[/tex] = Initial temperature = [tex]47^{\circ}\text{C}=(47+273.15)\ \text{K}[/tex]
[tex]P_2[/tex] = Final pressure = 3.87 atm
[tex]V_2[/tex] = Final volume = 197 mL
[tex]T_2[/tex] = Final temperature
We have the relation
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow T_2=\dfrac{P_2V_2}{P_1V_1}\times T_1\\\Rightarrow T_2=\dfrac{3.87\times 195}{2.77\times 263}\times (47+273.15)\\\Rightarrow T_2=331.63\ \text{K}=331.63-273.15\\\Rightarrow T_2=58.48^{\circ}\text{C}[/tex]
The final temperature will be [tex]58.48^{\circ}\text{C}[/tex].
