Question

Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 3.70 m×4.60 m×3.00m.

Answer 1

A lethal dose of HCN in air is about $L=300[\frac{{{\text{mg}}}}{{{\text{kg}}}}]$ (for dogs, see "O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 830")

We can use ideal gas law for air in the room under normal temperature and pressure condinitons (NTP, $T = 20[^\circ {\text{C}}] \approx 293[{\text{K}}]$ and $p = 1[{\text{atm}}] = 101.325 \cdot {10^3}[{\text{Pa}}]$)

$$pV = \nu RT = \frac{m}{M}RT$$

where $M = 28.97[\frac{{\text{g}}}{{{\text{mol}}}}] = 28.97 \cdot {10^{ - 3}}[\frac{{{\text{kg}}}}{{{\text{mol}}}}]$ - molar mass of dry air and $R = 8,31[\frac{{\text{J}}}{{{\text{K}} \cdot {\text{mol}}}}]$ - universal gas constant.

The volume of the room is $V = 3.70[{\text{m}}] \cdot 4.60[{\text{m}}] \cdot 3.00[{\text{m}}] = 51.06[{{\text{m}}^{\text{3}}}]$

Now we can get the mass of air in the room

$$m = \frac{{pV}}{{RT}}M = \frac{{101.325 \cdot {{10}^3}[{\text{Pa}}] \cdot 51.06[{{\text{m}}^{\text{3}}}]}}{{8.31[\frac{{\text{J}}}{{{\text{K}} \cdot {\text{mol}}}}] \cdot 293.15[{\text{K}}]}} \cdot 28.97 \cdot {10^{ - 3}}[\frac{{{\text{kg}}}}{{{\text{mol}}}}] = 61.53[{\text{kg}}]$$

Thus, the lethal amount of ${\text{HCN}}$ in this room will be

$${m_{HCN}} = L \cdot m = 300[\frac{{{\text{mg}}}}{{{\text{kg}}}}] \cdot 61.53[{\text{kg}}] = 18459[{\text{mg}}] \approx 18.46[{\text{g}}]$$