Question

I need help with 1,2,3, and 4

Answer 1

Answer:

• Problem 1: 1.85atm

• Problem 2: 110mL

• Problem 3: 290 mL

• Problem 4: 1.14 atm

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:

          $PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

         $P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

        $P_2=3.25atm\times755mL/1325mL=1.85atm$

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.

          $P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

           $P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

        $V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

        $V_2=110mL$

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:

         $V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

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:

        $\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

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.

• 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       $\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

    $865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$