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…

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 $\frac{v_1}{t_1}=\frac{v_2}{t_2}$. 2.19 l 3.33 l 3.68 l 5.60 l

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 $\frac{v_1}{t_1}=\frac{v_2}{t_2}$. 2.19 l 3.33 l 3.68 l 5.60 l

Answer

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 25.0 + 273.15=298.15\ K$ $T_2 = 40.0+ 273.15 = 313.15\ K$ $V_1 = 3.50\ L$

Step2: Rearrange the Charles's law formula

From $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we can get $V_2=\frac{V_1T_2}{T_1}$

Step3: Substitute values and calculate

$V_2=\frac{3.50\times313.15}{298.15}\approx3.68\ L$

Answer:

3.68 L