Answer
Write the equation of state:
$p V=m R T \ldots \ldots(1)$
Here, Absolute temperature of compressed air in tank is $T$, the mass of air $m$, Pressure of air in rigid tank is $p$, the gas constant is $R$
Since it is a rigid closed tank, the mass and the volume of the tank is constant. So, the equation
(1) becomes,
$\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}} \ldots \ldots$ (2)
Here, Absolute temperature of compressed air in tank is $T_{1}$, Final temperature with increase in pressure is $T_{2}$, initial pressure of air in the tank is $p_{1}$, final pressure of air in the tank is $p_{2}$
Substitute 90 psia for $p_{1}, 60^{\circ} \mathrm{F}$ for $T_{1}, 110^{\circ} \mathrm{F}$ for $T_{2}$ the values in equation (2).
$$
\begin{array}{l}
\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}} \\
\frac{90}{(60+459.67)}=\frac{p_{2}}{(110+459.67)} \\
p_{2}=\frac{(569.67)(90)}{(519.67)} \\
=98.65 \mathrm{psia}
\end{array}
$$
Calculate the increase in pressure.
$$
p_{\text {increase }}=p_{2}-p_{1}
$$
Substitute 90 psia for $p_{1}$ and $98.65$ psia for $p_{2}$.
$$
\begin{aligned}
p_{\text {incresse }} &=98.65-90 \\
&=8.65 \mathrm{psia}
\end{aligned}
$$
Therefore, the increase in pressure of air in the tank is $8.65 \mathrm{psia}$.
Work Step by Step
Write the equation of state:
$p V=m R T \ldots \ldots(1)$
Here, Absolute temperature of compressed air in tank is $T$, the mass of air $m$, Pressure of air in rigid tank is $p$, the gas constant is $R$
Since it is a rigid closed tank, the mass and the volume of the tank is constant. So, the equation
(1) becomes,
$\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}} \ldots \ldots$ (2)
Here, Absolute temperature of compressed air in tank is $T_{1}$, Final temperature with increase in pressure is $T_{2}$, initial pressure of air in the tank is $p_{1}$, final pressure of air in the tank is $p_{2}$
Substitute 90 psia for $p_{1}, 60^{\circ} \mathrm{F}$ for $T_{1}, 110^{\circ} \mathrm{F}$ for $T_{2}$ the values in equation (2).
$$
\begin{array}{l}
\frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}} \\
\frac{90}{(60+459.67)}=\frac{p_{2}}{(110+459.67)} \\
p_{2}=\frac{(569.67)(90)}{(519.67)} \\
=98.65 \mathrm{psia}
\end{array}
$$
Calculate the increase in pressure.
$$
p_{\text {increase }}=p_{2}-p_{1}
$$
Substitute 90 psia for $p_{1}$ and $98.65$ psia for $p_{2}$.
$$
\begin{aligned}
p_{\text {incresse }} &=98.65-90 \\
&=8.65 \mathrm{psia}
\end{aligned}
$$
Therefore, the increase in pressure of air in the tank is $8.65 \mathrm{psia}$.
