Answer
53.2 L
Work Step by Step
We find:
$n=89.2\,g\,CO_{2}\times\frac{1\,mol\,CO_{2}}{44.01\,g/mol}$
$=2.0268\,mol$
$P=737\,mmHg\times\frac{1\,atm}{760\, mmHg}$
$=0.970\,atm$
$T=(37+273)\,K=310\,K$
$V=\frac{nRT}{P}=\frac{(2.0268\,mol)(0.08206\,L\,atm\,K^{-1}mol^{-1})(310\,K)}{0.970\,atm}$
$=53.2\,L$
