Answer
$\Delta m_{added}=13.73lbm$
Work Step by Step
Knowing that:
$V_{1}=V_{2}$
From the ideal gas law:
$V_{1}=\frac{m_{1}RT_{1}}{P_{1}}=\frac{20lbm*0.3704\frac{psiaft^3}{lbmR}*(70+459.67)R}{20psia}=196.19ft^3$
$m_{2}=\frac{P_{2}V_{2}}{RT_{2}}=\frac{35psia*196.19ft^3}{0.3704\frac{psiaft^3}{lbmR}*(90+459.67)R}=33.73lbm$
Then the amount of air added to the tank is:
$\Delta m_{added}=33.73lbm-20lbm=13.73lbm$
