a hot hair balloon achieves its buoyant lift by heating the air inside the ballo

Question

a hot hair balloon achieves its buoyant lift by heating the air inside the balloon. Suppose the volume of a ballon is 1.8x10^3 m^3. The temperature of the air inside the balloon is 313K. and outside air temp is 274K. Assume the air is an ideal gas. Air density at 274K is 1.29 kg/m^3. a. how much weight can the balloon lift? b. show relation between density and temperature inside and outside of the balloon with no numbers, just derivation of formula. For this I got T1/T2= P1/P2 (density) but not sure if i'm right

Explanation / Answer

Answer: First, 2700 N lift is produced by removing that amount of WEIGHT from the balloon. So m=2700 N / 9.8 m/s2 = 275.5 kg. We have an equation n=PV/RT that tells us the number of moles in the balloon. But we need to know that air has mass 0.029 kg / mol before we can figure out how many moles must be removed from the balloon. n = 275.5 kg / (0.029 kg/mol) = 9483 mol So, we then have n2 – n1 = P2V2/RT2 - P1V1/RT1 But P is constant because the balloon is open to the atmosphere, and V is constant, so we end up with n2 – n1=(PV/R)(1/ T2 – 1/T1 ) = (PV/R)(1/ T2 – 1/273)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.