Answer
1.10 atm
Work Step by Step
$P_{1}=3.00\,atm$
$V_{1}=50.0\,mL$
$T_{1}= (8+273)K= 281\,K$
$T_{2}= (37+273)K=310\,K$
$V_{2}=150.0\,mL$
Combined gas law is
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (if n is constant)
$\implies P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}$
$=\frac{3.00\,atm\times50.0\,mL\times310\,K}{281\,K\times150.0\,mL}=1.10\,atm$
