Fox is at a party store getting balloons for his birthday. The attendant has Helium on tap in a metal canister. The canister has a volume of 8.0 m and is originally at a pressure of 2.20 x 10" Pa. The attendant uses 30 moles of Helium to fill 36 balloons. While filling the balloons, the air, canister, and balloons are all at room temperature, T=20 *C. (a) What is the pressure of the Helium in the canister after filling the balloons? (b) What is the volume of each balloon? You can assume that the air inside the balloons are at atmospheric pressure, 1.01 x105 Pa. (c) When Fox leaves the store, he steps outside into a chilly day. The cold weather draws heat from the balloons until the temperature of the air inside the balloons is 15 "C and they have shrunk somewhat. How much work is done by the surroundings to compress a single balloon? (d) How much heat has left each balloon? (Note: Helium is monatomic.) (e) What is the change in internal energy of a single balloon?(e) Change in internal energy of a single balloon: Assuming adiabatic conditions (AU 2 0): AU = Q W = 0 (7.74 x 103J) = 7.74 x 103J Therefore: (a) Pressure of Helium in the canister after filling the balloons is approximately 7.75 X 105 Pa. (b) Volume of each balloon is approximately 85.8 L. (0) Work done by the surroundings to compress a single balloon is approximately 7.74 x 103 J. (d) Heat leaving each balloon is approximately 7.74 x 103 J. (e) Change in internal energy of a single balloon is approximately 7.74 x 103 J. (b) Volume of each balloon: Using the ideal gas law for the air inside the balloons at atmospheric pressure: Patm X Vballoon = nRTroom Vballoon = nAlroom Patm Vballoon 36 X8.314 J/mol.K x 293 K = 1.01 x 105 Pa Vballoon ~85.8 L (c) Work done by the surroundings to compress a single balloon: Given that the atmospheric pressure is higher than the pressure inside the balloon, the work done by the surroundings to compress a single balloon is: W = -Patm X AV W = -1.01 x 10' Pa x (Vballoon - 8.0 m3) W~ -7.74 x 103 J (d) Heat leaving each balloon: Using the first law of thermodynamics Q = AU + W, where AU is the change in internal energy (assuming adiabatic conditions, no change in internal energy): Q =W = -7.74 x 103 JGiven: . Initial volume of the canister Vinitial = 8.0 m3 . Initial pressure of Helium Pinitial = 2.20 x 105 Pa . Number of moles of Helium used n = 30 moles . Room temperature Troom = 20 C = 293 K . Atmospheric pressure Patm = 1.01 x 105 Pa Given these values, let's solve each part: (a) Pressure of the Helium in the canister after filling the balloons: Using the ideal gas law PV = nRT: For Helium in the canister initially: Pinitial X Vinitial = nRTroom 2.20 x 10' Pa x 8.0m3 = 30 moles x 8.314J/mol . K x 293 K Solving gives: nremaining = 23.758 moles The pressure of Helium in the canister after filling the balloons: Pfinal = nremaining RI room Vinitial Pfinal = 23.758 moles x 8.314 J/mol. K x 293 K 8.0 m3 Pfinal ~ 7.75 x 10' Pa