Answer
$14.5\;\rm atm$
Work Step by Step
We have here a thermal interaction between the iron piece and the liquid nitrogen.
The nitrogen is at its boiling point which is at $-196^\circ$C and the iron piece is at room temperature which we can assume to be 20$^\circ $C.
And since the bottle is well-insulated, so there is no heat exchange with the surrounding which means that nitrogen gas will undergo an adiabatic process.
The heat lost by the iron piece will evaporate more nitrogen gas since the liquid nitrogen is at its boiling point.
Thus,
$$-Q_{iron}=Q_{niquid\;N_2}$$
The heat gained by the liquid nitrogen will convert more liquid into gas.
Thus,
$$-m_{ iron}c_{ iron}(T_f-T_{i,iron})=m_{1}L_{\rm v}$$
where $m_1$ is the mass of nitrogen gas that evaporated.
where $T_f$ is the boiling temperature of the nitrogen liquid. Thus,
$$-m_{iron}c_{ iron}(T_{liquid\;N_2}-T_{i,iron})=m_{1}L_{\rm v}$$
Thus,
$$m_1=\dfrac{-m_{iron}c_{ iron}(T_{liquid\;N_2}-T_{i,iron})}{L_{\rm v}}$$
Plugging the known;
$$m_1=\dfrac{ -(0.197)(449)(-196-20)}{ (1.99\times 10^5)}=\bf 0.09601\;\rm kg$$
$\bullet$ Now we know, before entering the iron piece, that the gas data was,
- $P_1=\bf 1\;\rm atm$
- $V_1=2000-500-25=\bf 1475\;\rm mL$
where the $25$mL is the escaped gas when the student entered the iron piece.
- $T_1=\bf -196^\circ\rm C$
So we can find the initial number of moles of nitrogen gas here by applying the ideal gas law.
$$P_1V_1=n_iRT_1$$
$$n_i=\dfrac{P_1V_1}{RT_1}=\dfrac{(1.013\times 10^5)(1475\times 10^{-3}\times 10^{-3})}{(8.31)(-196+273)}=\bf 0.2335\;\rm mol$$
$\bullet$ After some time of mixing the liquid iron into the liquid nitrogen, the gas data is,
- $T_2=\bf-196^\circ\rm C$
and now we need to find the total number of gas moles
$$n_f=n_i+n_1=n_i+\dfrac{m_1}{M_{N_2}} =n_i+\dfrac{m_1}{2M_{N}} $$
Plugging the known;
$$n_f=0.2335+\dfrac{0.09601\times 10^3}{2(14)}=\bf 3.662\;\rm mol $$
Now we need to find the final volume of the gas, so we have to find the volume of the evaporated gas
$$V_{m_1}=\dfrac{m_1}{\rho_{liquid\;N_2}}=\dfrac{0.09601}{810}=0.000118531\;\rm m^3=\bf 118.5\;\rm mL$$
Hence, the final volume of the gas is given by
$$V_2=V_1+V_{m_1}=1475+118.5=\bf 1593.5\;\rm mL$$
Now we can apply the ideal gas law to find the final pressure,
$$P_2=\dfrac{n_fRT_2}{V_2}=\dfrac{(3.662)(8.31)(-196+273)}{(1593.5\times 10^{-6})}=1.470\times 10^6\;\rm Pa$$
$$P_2=\color{red}{\bf 14.5}\;\rm atm$$
