Answer:
3) 670.7 °C (nearest tenth)
4) 818.2 torr (nearest tenth)
Explanation:
Question 3
Charles's Law
[tex]\sf \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\quad \textsf{(when pressure is constant and temperature is in kelvin)}[/tex]
where:
- [tex]\sf V_1[/tex] = first volume
- [tex]\sf V_2[/tex] = second volume
- [tex]\sf T_1[/tex] = first temperature
- [tex]\sf T_2[/tex] = second temperature
Given:
- [tex]\sf V_1[/tex] = 3 L
- [tex]\sf V_2[/tex] = 10 L
- [tex]\sf T_1[/tex] = 10 °C
First convert Celsius to kelvin:
Kelvin = Celsius + 273.15
⇒ [tex]\sf T_1[/tex] = 10 + 273.15 = 283.15 K
Substituting the given values into the formula:
[tex]\implies \sf \dfrac{3}{10}=\dfrac{283.15}{T_2}[/tex]
[tex]\implies \sf 3T_2=283.15 \cdot 10[/tex]
[tex]\implies \sf T_2=\dfrac{2831.5}{3}[/tex]
[tex]\implies \sf T_2=943.83333...\:K[/tex]
Converting kelvins back to Celsius:
[tex]\implies \sf T_2=943.83333...-273.15=670.68333...\:C^{\circ}[/tex]
Therefore, the final temperature will be 670.7 °C (nearest tenth).
Question 4
Gay-Lussac's Law
[tex]\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\quad \textsf{(when volume is constant and temperature is in kelvin)}[/tex]
where:
- [tex]\sf P_1[/tex] = first volume
- [tex]\sf P_2[/tex] = second volume
- [tex]\sf T_1[/tex] = first temperature
- [tex]\sf T_2[/tex] = second temperature
Given:
- [tex]\sf P_1[/tex] = 760 torr
- [tex]\sf T_1[/tex] = 27 °C
- [tex]\sf T_2[/tex] = 50 °C
First convert Celsius to kelvin:
kelvin = Celsius + 273.15
⇒ [tex]\sf T_1[/tex] = 27 + 273.15 = 300.15 K
⇒ [tex]\sf T_2[/tex] = 50 + 273.15 = 323.15 K
Substituting the given values into the formula:
[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=818.23754...\:torr[/tex]
Therefore, the new pressure is 818.2 torr (nearest tenth).
