Answer:
[tex]P_{gas}=131.96torr[/tex]
Explanation:
Helo,
In this case, the pressure must be computed as follows:
[tex]P_{gas}=gh\rho _{Hg}[/tex]
Which is using the density of mercury (13.6 g/mL) and its height, thus, we obtain (using the proper units):
[tex]P_{gas}=9.8\frac{m}{s^2}*(13.2cm *\frac{1m}{100cm} )*(13.6\frac{g}{cm^3}*\frac{1kg}{1000g}*\frac{1x10^6cm^3}{1m^3} )\\\\P_{gas}=17592.96Pa*\frac{760torr}{101325Pa} \\\\P_{gas}=131.96torr[/tex]
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