1. The problem statement, all variables and given/known data A well insulated 2 liter thermos bottle contains 500 ml of boiling liquid Nitrogen, and the remainder of the thermos has Nitrogen gas at 1 atmosphere in equilibrium with the liquid. You drop a 200 gram piece of iron at 25 degrees Celsius inside and seal the cap. After 5 seconds, what is the pressure inside the thermos? 2. Relevant equations [tex]\rho[/tex] liquid nitrogen = 810 kg/m3 [tex]\rho[/tex] iron = 7.87 g/cm3 PV = NkT I think that I need more equations to solve the problem, but I don't know what they are. 3. The attempt at a solution V total = 2 L = 0.002 m3 V liquid nitrogen = 500 ml = 500 * 10-3 L = 0.5 L = 0.0005 m3 before dropping the iron in: V nitrogen gas = 0.002 m3 - 0.0005 m3 = 0.0015 m3 V iron: [tex]\rho[/tex] = mV [tex]\rightarrow[/tex] 7.87 g/cm3 = 200g(V) [tex]\rightarrow[/tex] V = 0.03935 cm3 = 0.0003935 m3 after dropping the iron in: V nitrogen gas = 0.0015 m3 - 0.0003935 m3 = 0.0011065 m3 Tinitial of liquid nitrogen = boiling point at 1 atm = 77.2 K Tinitial of iron = 25 degrees Celsius = 298 K Pinitial of liquid nitrogen = Pinitial of nitrogen gas = 1 atm = 0.013 * 105 Pa [tex]\Delta[/tex]t = 5 seconds Pfinal = ? I just don't know which equation to use.