Question

A balloon originally had a volume of 4.39 L at 44C and a pressure of 729 torr  to what temperature must the balloon be cooled to reduce its volume to 3.78 L of the new pressure is at 1.0 atm

Answer 1

The new temperature : 11.56 °C

Further explanation

Boyle's law and Gay Lussac's law  

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

P1 = initial gas pressure (N/m² or Pa)  

V1 = initial gas volume (m³)  

P2 = final gas pressure  

V2 = final gas volume

T1 = initial gas temperature (K)  

T2 = final gas temperature  

V₁=4.39 L

T₁=44+273=317 K

P₁ = 729 torr = 0,959211 atm

V₂=3.78 L

P₂= 1 atm

$\tt \dfrac{0.959211\times 4.39}{317}=\dfrac{1\times 3.78}{T_2}\\\\T_2=\dfrac{1\times 3.78\times 317}{0.959211\times 4.39}\\\\T_2=284.559~K=11.56~C$