# A balloon that had a volume of 3.50 L at 25.0 °C is placed in a hot room at 40.0 °C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?

Jun 8, 2017

3.68 liters to three significant numbers.

#### Explanation:

This is a Charles Law problem

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

${V}_{1} = \text{3.5 L}$,

${T}_{1} = 25 + 273 = \text{298 K}$ ,

${V}_{2} = \text{unknown L}$,

${T}_{2} = 40 + 273 = \text{313 K}$,

so

$\left(\text{3.5 L")/("298 K") = V_2/("313 K}\right)$ .

This gives

$\left(\text{3.5 L") xx ("313 K")/("298 K}\right) = {V}_{2}$.

$\implies \text{3.68 L} = {V}_{2}$