Pressure increase in a chamber with 1 ml of water heated to 200 °C [closed]

Question

For safety reasons I want to know how much pressure will be generated in a sealed chamber with $1~\mathrm{ml}$ of water, pure nitrogen at $1.3~\mathrm{bar}$ at $22~\mathrm{^\circ C}$. This chamber will be heated to $200~\mathrm{^\circ C}$. What is the new pressure in the chamber? Chamber volume is $582~\mathrm{cm^3}$.

Answer 1

$1\ \mathrm{ml}$ of water at $p_0=1.3\ \mathrm{bar}$ and $T_0=22\ \mathrm{^\circ C}$ corresponds to a mass of $m_{\ce{H2O}}=0.99779\ \mathrm g$ and an amount of water of $n_{\ce{H2O}}=0.055386\ \mathrm{mol}$ (obviously not rounded to significant digits).

$582\ \mathrm{ml}-1\ \mathrm{ml}=581\ \mathrm{ml}$ of nitrogen at $p_0=1.3\ \mathrm{bar}$ and $T_0=22\ \mathrm{^\circ C}$ corresponds to a mass of $m_{\ce{N2}}=0.8624364\ \mathrm g$ and an amount of nitrogen of $n_{\ce{N2}}=0.030786609\ \mathrm{mol}$.

Now you know $T=200\ \mathrm{^\circ C}=473.15\ \mathrm K$, $V=582\ \mathrm{ml}$, and the amount of substance $n$ for water and nitrogen. Using the ideal gas law $pV=nRT$, you should get the partial pressures $p_{\ce{H2O}}=3.74\ \mathrm{bar}$ and $p_{\ce{N2}}=2.08\ \mathrm{bar}$. Thus, the total pressure is $p=5.82\ \mathrm{bar}$.

Better values can be obtained using so-called steam tables, which should give something like $p_{\ce{H2O}}=3.6711\ \mathrm{bar}$ and $p_{\ce{N2}}=2.0827\ \mathrm{bar}$. Therefore, the ideal gas law is a sufficient approximation for your purpose, especially considering the precision of your input data.