A balloon full of air has a volume of 2.75 L at a temperature of 255 K. What is the balloon's volume at 318 K?

Tuesday, March 11, 2014

A balloon full of air has a volume of 2.75 L at a temperature of 255 K. What is the balloon's volume at 318 K?

The relationship between the temperature and volume of a gas is described by Charles's Law:

$\displaystyle~\frac{{V}_{{1}}}{{T}_{{1}}}$ = $\displaystyle~\frac{{V}_{{2}}}{{T}_{{2}}}$

$\displaystyle~{V}_{{1}}$ = initial volume of the gas

$\displaystyle~{T}_{{1}}$ = initial temperature of the gas

$\displaystyle~{V}_{{2}}$ = final volume of the gas

$\displaystyle~{T}_{{2}}$ = final temperature of the gas

(Assume that the pressure and number of moles remain constant.)

There is an direct relationship between the temperature and volume of a gas. When the temperature of a gas increases, the volume also increases. The reverse is also true - when the temperature of a gas decreases, the volume also decreases.

According to the problem,

$\displaystyle~{V}_{{1}}$ = 2.75 L

$\displaystyle~{T}_{{1}}$ = 255 K

$\displaystyle~{V}_{{2}}$ = ? L

$\displaystyle~{T}_{{2}}$ = 318 K

Step 1: Rearrange the formula for Charles's Law to isolate $\displaystyle~{V}_{{2}}$ .

$\displaystyle~{V}_{{2}}$ = $\displaystyle~{V}_{{1}}{T}_{{2}}$ $\displaystyle\div$ $\displaystyle~{T}_{{1}}$

Step 2: Plug in the given values and calculate.

$\displaystyle~{V}_{{2}}$ = [(2.75 L)(318 K)]/255 K = 3.43 L

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Tuesday, March 11, 2014