Answer
$1.93\times10^{5}\,Pa$
Work Step by Step
We find:
$PV=nRT$ (Ideal gas law)
$V= 2.10\,mL=2.10\,mL\times\frac{10^{-6}\,m^{3}}{1\,mL}=2.10\times10^{-6}\,m^{3}$
$n=6.15\times10^{-3}g\times\frac{1\,mol\,CO_{2}}{44.01\,g}=1.3974\times10^{-4}\,mol$
$T=(75+273.15)\,K=348.15\,K$
$R=8.314472\,Jmol^{-1}K^{-1}$
Then, $P=\frac{nRT}{V}=\frac{(1.3974\times10^{-4}\,mol)(8.314472\,Jmol^{-1}K^{-1})(348.15\,K)}{2.10\times10^{-6}\,m^{3}}$
$=1.93\times10^{5}\,Pa$
