Answer:
[tex] \qquad \large \bold{818.2 \: torr \: (nearest \: tenth)} \\ [/tex]
Explanation:
[tex] \\ \qquad\qquad \large \underline {\sf{ \blue{\pmb{Gay-Lussac's \: Law : }}}} \\ [/tex]
[tex]\dashrightarrow\:\sf{\dfrac{P_1}{T_1}=\dfrac{P_2}{T_1}}\begin{aligned} \: \\ \\\textsf{(when volume is constant and temperature is in kelvin)} \end{aligned}\\ [/tex]
[tex] \sf{where:} \\ [/tex]
- [tex]\sf{ P_1 = \underline{ \pmb{ first \: volume}}} \\ [/tex]
- [tex]\sf{ P_2 = \underline{ \pmb{ second \: volume}}} \\ [/tex]
- [tex]\sf T_1 = \underline{ \pmb{ first \: temperature}} \\ [/tex]
- [tex]\sf{ T_2 = \underline{ \pmb{second \: temperature}}} \\ [/tex]
[tex] \sf{Given:} \\ [/tex]
- [tex]\bold{P_1 = \underline{ \pmb{760 \: torr}}} \\ [/tex]
- [tex]\bold{T_1 = \underline{ \pmb{27 ^{\circ}C}}} \\ [/tex]
- [tex]\bold{T_2 = \underline{ \pmb{ 50 ^{\circ}C} }}\\ [/tex]
[tex] \underline{ \sf{ \pmb{First \: convert \: Celsius \: to \: kelvin:}}} \\ [/tex]
[tex] \qquad \underline{\sf{ \: \: kelvin = Celsius + 273.15 \: \: } }\\ [/tex]
- [tex]\bold{ T_1 = 27 + 273.15 = \underline{ \pmb{ 300.15 \: K}}} \\ [/tex]
- [tex] \bold{ T_2 = 50 + 273.15 = \underline{ \pmb{323.15 \: K }}}\\ [/tex]
[tex] \sf{ \pmb{Substituting} \: the \: \pmb{given \: values }\: into \: the \: \pmb{ formula}:} \\ [/tex]
[tex]\implies \: \sf \dfrac{760}{300.15}=\dfrac{P_2}{323.15} \\ [/tex]
[tex]\implies \: \sf 760 \cdot 323.15=P_2 \cdot 300.15 \\ [/tex]
[tex]\implies \: \sf 245594=300.15\:P_2 \\ [/tex]
[tex]\implies \: \sf P_2=\dfrac{245594}{300.15} \\ [/tex]
[tex]\implies \: \sf P_2= \pmb{818.23754. . . \:torr} \\ \\ [/tex]
Therefore, the new pressure is 818.2 torr (nearest tenth).
