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?

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?

Avery Adams · Accepted Answer

3.68 liters to three significant numbers.

This issue pertains to Charles Law. # V_1/T_1 = V_2/T_2# # V_1 = "3.5 L"#, # T_1 = 25 + 273 = "298 K"# , #V_2 = "unknown L"#, # T_2 = 40 + 273 = "313 K"#, thus # ("3.5 L")/("298 K") = V_2/("313 K")# . This provides # ("3.5 L") xx ("313 K")/("298 K") = V_2#. #=> "3.68 L" = V_2#

Axel Alvarado · Answer

undefined You can use the combined gas law to solve this problem. The combined gas law equation is:

$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $

Given:
$V_1 = 3.50 \, \text{L}$
$T_1 = 25.0 + 273.15 \, \text{K}$
$T_2 = 40.0 + 273.15 \, \text{K}$
$P_1 = P_2 = 1.00 \, \text{atm}$

Substituting the values into the equation and solving for $V_2$:

$ \frac{(1.00 \, \text{atm})(3.50 \, \text{L})}{(25.0 + 273.15) \, \text{K}} = \frac{(1.00 \, \text{atm})(V_2)}{(40.0 + 273.15) \, \text{K}} $

$ V_2 = \frac{(1.00 \, \text{atm})(3.50 \, \text{L})(313.15 \, \text{K})}{(298.15 \, \text{K})} $

$ V_2 ≈ 3.77 \, \text{L} $

Therefore, the new volume of the balloon in the hot room is approximately 3.77 L.