# Pressure inside a thermos

1. Oct 6, 2008

### spherical23

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

$$\rho$$ liquid nitrogen = 810 kg/m3

$$\rho$$ 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: $$\rho$$ = mV $$\rightarrow$$ 7.87 g/cm3 = 200g(V) $$\rightarrow$$ 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

$$\Delta$$t = 5 seconds

Pfinal = ?

I just don't know which equation to use.

2. Oct 6, 2008

### mgb_phys

You need to work out how much heat energy the iron gives up when cooled (assume final T of iron = LN2)
Assume this energy goes into boiling extra LN2.
Work out how much extra N2 this generates.
Then it's just gas laws for the pressure.

The 5 seconds is tricky, are you sure about that part? Unless you are expected to calcualte heat diffusion rate from the iron it's irrelevant.

3. Oct 6, 2008

### spherical23

thanks, i will give that a try and then reply again when i come up with something. i think the 5 seconds is just enough for relaxation time so that we don't have to worry about anything during the relaxation time.