Certa quantidade de um gás ideal ocupa um volume de 30 litros à temperatura de 77 °C e sob pressão de 2,0 atmosferas. Responda:a) Qual o volume dessa

\[\dfrac{{{{\text{P}}_1} \cdot {{\text{V}}_1}}}{{{{\text{T}}_1}}} = \dfrac{{{{\text{P}}_2} \cdot {{\text{V}}_2}}}{{{{\text{T}}_2}}}\]

\[\eqalign{ & {\text{Onde:}} \cr & {\text{P = Pressão}} \cr & {\text{V = Volume}} \cr & {\text{T = Temperatura (em Kelvin}}{\text{, K)}} }\]

Temos que transformar as temperaturas de célsius para kelvin:

\[\eqalign{ {{\text{T}}_1} &= {\text{ célsius + 273}}\cr{{\text{T}}_1} &= {\text{ 77 + 273}}\cr{{\text{T}}_1} &= {\text{ 350 K}} }\]

\[\eqalign{ {{\text{T}}_1} &= {\text{ célsius + 273}}\cr{{\text{T}}_1} &= {\text{ 427 + 273}}\cr{{\text{T}}_1} &= {\text{ 700 K}} }\]

a)

Para encontrar a resposta basta substituir na fórmula:

\[\eqalign{ \dfrac{{{{\text{P}}_1} \cdot {{\text{V}}_1}}}{{{{\text{T}}_1}}} &= \dfrac{{{{\text{P}}_2} \cdot {{\text{V}}_2}}}{{{{\text{T}}_2}}}\cr\dfrac{{2 \cdot 30}}{{350}} &= \dfrac{{8 \cdot {{\text{V}}_2}}}{{700}}\cr\dfrac{{60}}{{350}} \cdot 700 &= 8 \cdot {{\text{V}}_2}\cr120 &= 8 \cdot {{\text{V}}_2}\cr{{\text{V}}_2} &= \dfrac{{120}}{8}\cr{{\text{V}}_2} &= 15{\text{ L}} }\]

Concluímos então que \(\boxed{{{\text{V}}_2} = 15{\text{ L}}}\).

b)

\[\eqalign{ \dfrac{{{{\text{P}}_1} \cdot {{\text{V}}_1}}}{{{{\text{T}}_1}}} &= \dfrac{{{{\text{P}}_2} \cdot {{\text{V}}_2}}}{{{{\text{T}}_2}}}\cr\dfrac{{2 \cdot 30}}{{350}} &= \dfrac{{2 \cdot 120}}{{{{\text{T}}_2}}}\cr0,17 \cdot {{\text{T}}_2} &= 240\cr{{\text{T}}_2} &= \dfrac{{240}}{{0,17}}\cr{{\text{T}}_2} &= 1.411,76{\text{ K}} }\]

Transformando para célsius:

\[\eqalign{ {{\text{T}}_2} &= {\text{kelvin}} - 273\cr{{\text{T}}_2} &= 1.411,76 - 273\cr{{\text{T}}_2} &= 1.138,76{\text{ °C}} }\]

Concluímos que a temperatura será 1.138,76 °C.