Answer:
Explanation:
Problem 1
1. Data
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
2. Formula
Since the temeperature is constant you can use Boyle's law for idial gases:
[tex]PV=constant\\\\P_1V_1=P_2V_2[/tex]
3. Solution
Solve, substitute and compute:
[tex]P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2[/tex]
[tex]P_2=3.25atm\times755mL/1325mL=1.85atm[/tex]
Problem 2
1. Data
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
2. Formula
You assume that the temperature does not change, and then can use Boyl'es law again.
[tex]P_1V_1=P_2V_2[/tex]
3. Solution
This time, solve for V₂:
[tex]P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2[/tex]
Substitute and compute:
[tex]V_2=548mmHg\times 125mL/625mmHg=109.6mL[/tex]
You must round to 3 significant figures:
[tex]V_2=110mL[/tex]
Problem 3
1. Data
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
2. Formula
At constant pressure, Charle's law states that volume and temperature are inversely related:
[tex]V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]
The temperatures must be in absolute scale.
3. Solution
a) Convert the temperatures to kelvins:
- T₁ = 25 + 273.15K = 298.15K
- T₂ = 35 + 273.15K = 308.15K
b) Substitute in the formula, solve for V₂, and compute:
[tex]\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml[/tex]
You must round to two significant figures: 290 ml
Problem 4
1. Data
a) P = 865mmHg
b) Convert to atm
2. Formula
You must use a conversion factor.
Divide both sides by 760 mmHg
[tex]\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}[/tex]
3. Solution
Multiply 865 mmHg by the conversion factor:
[tex]865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer[/tex]
