A gas evolved during the fermentation of glucose (wine making) has a volume of \(0.78 \mathrm{~L}\) at \(20.1^{\circ} \mathrm{C}\) and 1.00 atm. What was the volume of this gas at the fermentation temperature of \(36.5^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) pressure?

Short Answer

Expert verified
The final volume of the gas at the fermentation temperature of \( 36.5^\circ \mathrm{C} \) and \( 1.00 \mathrm{~atm} \) pressure is approximately \( 0.82 \mathrm{~L} \)

Step by step solution

01

Understand Charles's Law and Its Formula

Charles's Law states that the volume of a gas (V) is directly proportional to its absolute temperature (T) when the pressure is held constant. The general formula is \( V1/T1 = V2/T2 \), where V1 and T1 are initial volume and temperature, and V2 and T2 are final volume and temperature.
02

Convert temperatures to Kelvin

In Kelvin, initial temperature T1 is \(20.1^\circ C+273.15 = 293.25 \mathrm{~K} \) and final temperature T2 is \(36.5^\circ C+273.15 = 309.65 \mathrm{~K} \)
03

Substitute the given data into the formula

Substitute the initial volume \( V1=0.78 \mathrm{~L} \), initial temperature \( T1=293.25 \mathrm{~K} \), and final temperature \( T2=309.65 \mathrm{~K} \) into the formula.
04

Solve for the final volume

To find the unknown volume \( V2 \), re-arrange the formula: \( V2 = V1 \times T2/T1 \). And substitute the given value in.

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