Question

A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

Answer 1

Data:

`V_(initial) = 590\:mL`

`T_(initial) = -55.0^0C`
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 →
`T_(initial) = 218.0\:K`

`V_(final) = ? (in\:milliliters)`

`T_(final) = 30.0^0C`
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 →
`T_(final) = 303.0\:K`

By the first Law of Charles and Gay-Lussac, we have:

`( V_(i) )/( T_(i) ) = ( V_(f) )/( T_(f) )`

Solving:

`( V_(i) )/( T_(i) ) = ( V_(f) )/( T_(f) )`

`( 590 )/( 218.0 ) = ( V_(f) )/( 303.0 )`
Product of extremes equals product of means:

`218.0* V_(f) = 590*303.0`

`218.0 V_(f) = 178770`

`V_(f) = (178770)/(218.0)`

`\boxed{\boxed{V_(f) \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark`