Question

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? Use mc005-1.jpg.

Answer 1

By using charles law

$\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$ Pressure is constant

In the given question

V₁ = 3.5 L
T₁ = 25°C = (25+273) K = 298 K

V₂ = ?
T₂ = 40°C = (40+273) K = 313 K

$V_{2} = \frac{V_{1}T_{2}}{T_{1}} = \frac{3.5 \times313}{298} = 3.676 L$