Question

What volume did a helium-filled balloon have at 18.1 °C and 2.61 atm if its new volume was 59.9 mL at 1.92 atm and 12.5°C?

Answer 1

Answer:

a.  V=44.9mL

Explanation:

For a gas undergoing all of these changes, it will be important to combine Boyle's Law and Charles' Law to form the following equation (if it isn't already known):

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

Where the Temperatures must be measured in Kelvin.

Recall that to convert Celsius to Kelvin, one must add 273 or use the equation $T_C+273=T_K$.

Thus, $T_1=(18.1+273)[K]=291.1[K]$  and $T_2=(12.5+273)[K]=285.5[K]$

To solve for the requested quantity, note that all of the other units match between beginning and end, so we substitute and solve:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$\dfrac{(2.61[atm])V_1}{(291.1[K])}=\dfrac{(1.92[atm])(59.9[mL])}{(285.5[K])}$

$\dfrac{(2.61[atm] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})\bold{V_1}}{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}*\dfrac{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}=\dfrac{(1.92[atm]\!\!\!\!\!\!\!\!\!\!\!{--})(59.9[mL])}{285.5[K\!\!\!\!\!{-}]}*\dfrac{291.1[K\!\!\!\!\!{-}]}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!{--}}$

$V_1=44.928677576[mL]$

Accounting for significant digits, $V_1=44.9[mL]$