Question

Can someone help me here please​

Answer 1

Answer:

3)  670.7 °C (nearest tenth)

4)  818.2 torr (nearest tenth)

Explanation:

Question 3

Charles's Law

$\sf \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\quad \textsf{(when pressure is constant and temperature is in kelvin)}$

where:

• $\sf V_1$ = first volume
• $\sf V_2$ = second volume
• $\sf T_1$ = first temperature
• $\sf T_2$ = second temperature

Given:

• $\sf V_1$ = 3 L
• $\sf V_2$ = 10 L
• $\sf T_1$ = 10 °C

First convert Celsius to kelvin:

Kelvin = Celsius + 273.15

⇒ $\sf T_1$ = 10 + 273.15 = 283.15 K

Substituting the given values into the formula:

$\implies \sf \dfrac{3}{10}=\dfrac{283.15}{T_2}$

$\implies \sf 3T_2=283.15 \cdot 10$

$\implies \sf T_2=\dfrac{2831.5}{3}$

$\implies \sf T_2=943.83333...\:K$

Converting kelvins back to Celsius:

$\implies \sf T_2=943.83333...-273.15=670.68333...\:C^{\circ}$

Therefore, the final temperature will be 670.7 °C (nearest tenth).

Question 4

Gay-Lussac's Law

$\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\quad \textsf{(when volume is constant and temperature is in kelvin)}$

where:

• $\sf P_1$ = first volume
• $\sf P_2$ = second volume
• $\sf T_1$ = first temperature
• $\sf T_2$ = second temperature

Given:

• $\sf P_1$ = 760 torr
• $\sf T_1$ = 27 °C
• $\sf T_2$ = 50 °C

First convert Celsius to kelvin:

kelvin = Celsius + 273.15

⇒ $\sf T_1$ = 27 + 273.15 = 300.15 K

⇒ $\sf T_2$ = 50 + 273.15 = 323.15 K

Substituting the given values into the formula:

$\implies \sf \dfrac{760}{300.15}=\dfrac{P_2}{323.15}$

$\implies \sf 760 \cdot 323.15=P_2 \cdot 300.15$

$\implies \sf 245594=300.15\:P_2$

$\implies \sf P_2=\dfrac{245594}{300.15}$

$\implies \sf P_2=818.23754...\:torr$

Therefore, the new pressure is 818.2 torr (nearest tenth).

Answer 2

Answer:

3. 670.3 °C

4. 807.5 torr

Explanation:

Solving :

Q3

• V₁/T₁ = V₂/T₂
• 3 / 283 = 10 / T₂
• T₂ = 2830/3
• T₂ = 943.3 K
• T₂ = 670.3 °C

Q4

• P₁/T₁ = P₂/T₂
• 760/300 = P₂/323
• P₂ = 760/300 x 323
• P₂ = 2.5 x 323
• P₂ = 807.5 torr