Question

A sample of gas occupies 54.0 mL at 24.0C. What is the volume of the gas at 80.0C?

Answer 1

The volume of the sample of gas at 80 °C is 64.18 mL.

From the question given above, the following data were obtained:

Initial volume (V₁) = 54 mL

Initial temperature (T₁) = 24 °C = 24 + 273 = 297 K

Final temperature (T₂) = 80 °C = 80 + 273 = 353 K

Final volume (V₂) =?

Using the Charles' law equation, the new volume of the gas can be obtained as follow:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} \\\\\frac{54}{297} = \frac{V_{2}}{353}$

Cross multiply

297 × V₂ = 54 × 353

297 × V₂ = 19062

Divide both side by 297

$V_{2} = \frac{19062}{297}$

V₂ = 64.18 mL

Therefore, the volume of the gas at 80 °C is 64.18 mL.

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Answer 2

Answer:

Your answer is  64.18 mL.

Explanation:

First you cross multiply

297 × V₂ = 54 × 353

297 × V₂ = 19062

You then divide both sides by 297 when will the give you

V₂ = 64.18 mL