Question

Question 1 (3 points)
A sample of xenon gas has a volume of 6.33 L at a temperature of 35.0 degrees celsius and a pressure of 342 mmHg. What would the volume
be if the temperature rose to 57.9 degrees celsius and the pressure Increased to 821 mmHg?
Volume =
Liters

Answer 1

Answer:

Explanation:

ΔTemp => 35⁰C(108K) increases to 57.9⁰C(330.9L) => increases volume (Charles Law)

Use the Kelvin Temperature values in a ratio that will increase the original volume.

ΔVol = 6.33L(330.9/108.0) => gives a larger volume. Using 108.0/330.9 would give a smaller volume and would be contrary to what the problem is asking.

ΔPress => 342 mmHg increases to 821 mmHg => decreases volume (Boyles Law)

Use the pressure values in a ratio that will decrease the original volume.

ΔPress = 6.33L(342/821) => gives a smaller volume. Using 821/342 would give a larger volume and would be contrary to what the problem is asking.

Now, putting both ΔTemp together with ΔPress => net change in volume. (Combined Gas Law)

ΔVol = 6.33L(330.9/108.0)(342/821) = 8.08L (final volume of gas).

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This problem can also be worked using the combined gas law equation:

P₁V₁/T₁ = P₂V₂/T₂ => V₂ = P₁V₁T₂/T₁P₂

V₂ = [(342mm)(6.33L)(330.9K)]/[(108K)(821mm)] = 8.08L (final volume of gas)