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

Answer:

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

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:

$\displaystyle \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$  [Charles Law]

Solution:

Put the givens

$\displaystyle \frac{7}{573 } =\frac{V_2}{298} \\\\0.01 = \frac{V_2}{298} \\\\\times 298 \ to \ both \ sides\\\\0.01 \times 298 = V_2\\\\3.64 \ L =V_2\\\\\boxed{V_2 =3.64 \ L}\\\\\rule[225]{225}{2}$