Question

A balloon at 30.0°C has a volume of 222 mL. If the temperature is increased to 56.1°C and the pressure remains constant, what will the new volume be, in mL?

Answer 1

Answer:

V₂ ≈ 241 mL

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Equality Properties

Chemistry

Unit 0

• Temperature Conversion: K = °C + 273.15

Gas Laws

Charles' Law: $\displaystyle \frac{V_1}{T_1} = \frac{V_2}{T_2}$

• V is volume
• T is temperature (in Kelvins)

Explanation:

Step 1: Define

V₁ = 222 mL

T₁ = 30.0 °C

T₂ = 56.1 °C

Step 2: Identify Conversions

Temp Conversion

Step 3: Convert

T₁ - 30.0 °C + 273.15 = 303.15 K

T₂ - 56.1 °C + 273.15 = 329.25 K

Step 4: Solve for V₂

1. Substitute [CL]:                    $\displaystyle \frac{222 \ mL}{303.15 \ K} = \frac{V_2}{329.25 \ K}$
2. Cross-multiply:                    $\displaystyle 73093.5 \ mL \cdot K = 303.15(V_2) \ K$
3. Isolate V₂:                            $\displaystyle 241.113 \ mL = V_2$
4. Rewrite:                               $\displaystyle V_2 = 241.113 \ mL$

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

241.113 mL ≈ 241 mL

Answer 2

The new volume of the balloon will be 241.12 mL

Data obtained from the question

•Initial temperature (T₁) = 30 °C = 30 + 273 = 303 K

•Initial Volume (V₁) = 222 mL

•Pressure = Constant

•Final temperature (T₂) = 56.1 °C = 56.1 + 273 = 329.1

•Final volume (V₂) =?

The final volume (i.e new volume) of the balloon can be obtained by using the Charles' law equation as shown below:

V₁ / T₁ = V₂ / T₂

222 / 303 = V₂ / 329.1

Cross multiply

303 × V₂ = 222 × 329.1

303 × V₂ = 73060.2

Divide both side by 303

V₂ = 73060.2 / 303

V₂ = 241.12 mL

Learn more about gas laws:

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