 # A rigid tank contains air at a pressure of 90 dedica66em 2021-12-17 Answered
A rigid tank contains air at a pressure of 90 psia and a temperature of ${60}^{\circ }F$ . By how much will the pressure increase as the temperature is increased to ${110}^{\circ }{F}^{2}$
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Given Information:
$P1=90PSITI=60F=\left(60+459.67\right)R$
$P2=????T2=110F=\left(110+459.67\right)R$
$p=\rho RT$
Since V is constant and mass does not change, density is constant. R is also constant The equation can be simplified to the pressure law.
$P2=P1\left(\frac{T2}{T1}\right)=98.67$

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$p=\rho RT$ (Eq.1.8)
For a riqid closed tank the air mass and olume are constant so p=constant. Thus, from Eq.1.8 (with R constant)
$\frac{{p}_{1}}{{T}_{1}}=\frac{{p}_{2}}{{T}_{2}}$ (1)
where ${p}_{1}=90\psi a$, ${T}_{1}=60°F+460=520°R$,
and ${T}_{2}=110°F+460=570°R$ From Eq.(1)
${p}_{2}=\frac{{T}_{2}}{{T}_{1}}{p}_{1}=\left(\frac{570°R}{520°R}\right)\left(90\psi a\right)=98.7\psi a$

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