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 (P_1.V_1)/(T_1)=(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 (0.959211* 4.39)/(317)=(1* 3.78)/(T_2)\\\\T_2=(1* 3.78* 317)/(0.959211* 4.39)\\\\T_2=284.559~K=11.56~C`