Question

A balloon inflated in a room at 300 ºC has a volume of 7.0 L. The balloon is then cooled to a temperature of 25.0 ºC. What is the new volume if the pressure remains constant?

Answer 1

`\huge\boxed{\sf V_2 =3.64 \ L}`

Step-by-step explanation:

Given Data:

Initial Volume =
`V_(1)` = 7.0 L

Initial Temperature =
`T_(2)` = 300 °C + 273 = 573 K

Final Temperature =
`T_(2)` = 25.0 °C + 273 = 298 K

Required:

Final Volume =
`V_(2)` = ?

Formula:

`\displaystyle (V_(1))/(T_(1)) = (V_(2))/(T_(2))` [Charles Law]

Solution:

Put the givens

`\displaystyle (7)/(573 ) =(V_2)/(298) \\\\0.01 = (V_2)/(298) \\\\* 298 \ to \ both \ sides\\\\0.01 * 298 = V_2\\\\3.64 \ L =V_2\\\\\boxed{V_2 =3.64 \ L}\\\\\rule[225]{225}{2}`