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?

a. 44.9
b. 63.8
c.43.2
d. 83.0

Answer 1

a. V=44.9mL

Step-by-step 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):

`(P_1V_1)/(T_1)=(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:

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

`((2.61[atm])V_1)/((291.1[K]))=((1.92[atm])(59.9[mL]))/((285.5[K]))`

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

`V_1=44.928677576[mL]`

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