You come into the lab one day and find a well-insulated $2000\mathrm{mL}$ thermos bottle containing $500\mathrm{mL}$ of boiling liquid nitrogen. The remainder of the thermos has nitrogen gas at a pressure of $1.0$ atm. The gas and liquid are in thermal equilibrium. While waiting for lab to start, you notice a piece of iron on the table with " $197\mathrm{g}$" written on it. Just for fun, you drop the iron into the thermos and seal the cap tightly so that no gas can escape. After a few seconds have passed, what is the pressure inside the thermos? The density of liquid nitrogen is $810\mathrm{kg}/{\mathrm{m}}^3$.

Well-insulated thermos bottle $=2000mL$

Boiling liquid nitrogen $=500mL$

Pressure$=1.0atm$

The density of liquid nitrogen is$810kg/m^3$

Let us first make the situation clear: the iron will give away heat to the environment until they are in thermal equilibrium This means that, taking the boiling point of the nitrogen and the initial temperature of the iron as our temperature limits, there are two possibilities:

• The iron has less heat than the heat of evaporation of the entire liquid nitrogen, meaning that all the iron will cool down to the boiling point of the nitrogen and not all nitrogen will have evaporated, or,
• The iron has more heat than the heat needed to evaporate all the liquid nitrogen. In this case, the iron will cool down until all the liquid nitrogen has evaporated and then further cool down until it reaches thermal equilibrium with the nitrogen gas.

As you can imagine, it is much easier to find out which of these cases we have than to attempt a totally parametric solution.

The $500\mathrm{ml}$contains the following mass of liquid nitrogen:

localid="1648621202237" role="math" $m=\rho V=\left(810kg/m^3\times 5\times {10}^{-4}{m}^3\right)=0.405\mathrm{kg}$

If from room temperature (we'll have to assume it, let it be $25°\mathrm{C}$) our iron is cooled until the boiling point of liquid nitrogen it will give the following heat

localid="1648621410902" role="math" $Q=c{m}_i\Delta T=\left(449\times 0.197g\right)(25K-(-196K\left)\right)=19550J.$

Now, knowing the nitrogen latent heat of evaporation, let us calculate the heat needed to evaporate the entire mass of liquid nitrogen:

$Q=mL=\left(0.405kg\times 1.99·{10}^{5}J/kg\right)=\underset{¯}{80595\mathrm{J}}$

This means that the first case is what we have. Therefore we will calculate the amount of liquid nitrogen that the heat from the iron could evaporate. This could also be done using the ratio of heats that we have, but for the sake of reminding the formulas, we'll write

$m_e=\frac{Q}{L}=\left(\frac{19550J}{1.99·{10}^{5}J/kg}\right)=\underset{¯}{0.09824\mathrm{kg}}$

As we'll see, it's easier to have this amount in moles. We can easily switch knowing that the molar mass of diatomic nitrogen is $28\mathrm{g}/\mathrm{mol}$.

localid="1648621478106" role="math" $n=\frac{m}{M}=\left(\frac{98.24kg}{28g/mol}\right)=\underset{¯}{3.5086\mathrm{mol}}$

So, what we will have is that the amount of gas (mole number) will increase, but the volume will increase as well, since the evaporated liquid nitrogen had a significant volume. Therefore we'll have to take this into account as well (although as you'll see, the book hasn't).

Knowing the mass that evaporated, $m_e$, the volume that was "freed up" is,

localid="1648621512722" $V=\frac{m_e}{\rho }=\left(\frac{0.09824kg}{810kg/m^3}\right)=1.2128·{10}^{-4}{\mathrm{m}}^3$

This means that the new volume of nitrogen gas will be

localid="1648621532613" $V_2=V_1+V=\left(1.500m^3+0.121m^3\right)=1.621\text{liters}$

Please do remember that since the iron was cooled down to the liquid nitrogen boiling temperature, the temperature of the entire system after this cooling down is the same as the previous temperature, the boiling point of liquid nitrogen.

From the Ideal Gas Law we have

$pV=nRT\Rightarrow \frac{pV}{n}=RT$

The reason we did this separation is that the $RT$in the left, as we said, remains a constant. Therefore, for the initial state (l) and final state (2) we have,

$\frac{p_1{V}_1}{n_1}=\frac{p_2{V}_2}{n_2}$

The only thing we are left to find is initial number of moles. This can be done by applying the Ideal Gas Law in the initial state:

localid="1648621589168" role="math" $p_1{V}_1=n_{1}RT\Rightarrow n_1=\frac{p_1{V}_1}{RT}=\left(\frac{1.01\times {10}^5{m}^3\times 0.0015m^3}{8.314\times (273K-196K)}\right)=\underset{¯}{0.2367\mathrm{moles}}$

Therefore, considering that we have to add the number of moles, our final pressure will be

localid="1648621651224" $p_2=\left(\frac{3.5086mol+0.2367mol}{0.2367mol}\times \frac{1.5mol}{1.621mol}\times 1\mathrm{atm}\right)$

Multiply the values,

$=14.64$

$\approx 15\mathrm{atm}$

Hence, the pressure inside the thermos is$15atm.$