Answer
$\Delta m=0.026lbm$
Work Step by Step
$P_{abs,1}=14.6psia+20psig=34.6psia$
$P_{abs,2}=14.6psia+30psig=44.6psia$
From the ideal gas law:
$m=\frac{PV}{RT}$
$m_{1}=\frac{34.6psia*0.53ft^3}{0.3704\frac{psiaft^3}{lbmR}*(90+459.67)R}=0.090lbm$
$m_{2}=\frac{44.6psia*0.53ft^3}{0.3704\frac{psiaft^3}{lbmR}*(90+459.67)R}=0.116lbm$
Then:
$\Delta m=0.116lbm-0.090lbm=0.026lbm$
