Question

Can someone help me here please​

Answer 1

3) 670.7 °C (nearest tenth)

4) 818.2 torr (nearest tenth)

Step-by-step explanation:

Question 3

Charles's Law

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

where:

• 
  `\sf V_1` = initial volume
• 
  `\sf V_2` = final volume
• 
  `\sf T_1` = initial temperature
• 
  `\sf T_2` = final temperature

Given values:

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

Convert the temperature in Celsius to kelvin by adding 273.15:

⇒
`\sf T_1` = 10 + 273.15 = 283.15 K

Substitute the given values into the formula and solve for T₂:

`\implies \sf (3)/(10)=(283.15)/(T_2)`

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

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

where:

• 
  `\sf P_1` = initial pressure
• 
  `\sf P_2` = final pressure
• 
  `\sf T_1` = initial temperature
• 
  `\sf T_2` = final temperature

Given values:

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

Convert the temperature in Celsius to kelvin by adding 273.15:

⇒
`\sf T_1` = 27 + 273.15 = 300.15 K

⇒
`\sf T_2` = 50 + 273.15 = 323.15 K

Substitute the given values into the formula and solve for P₂:

`\implies \sf (760)/(300.15)=(P_2)/(323.15)`

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

`\implies \sf 245594=300.15 \cdot P_2`

`\implies \sf P_2=(245594)/(300.15)`

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

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