Step by step solution to "A rigid tank contains air at a pressure..."

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dedica66em

Answered question

2021-12-17

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}

Answer & Explanation

Jacob Homer

Beginner2021-12-18Added 41 answers

Given:
$P 1 = 90 P S I T I = 60 F = (60 + 459.67) R$
$P 2 = ? ? ? ? T 2 = 110 F = (110 + 459.67) R$
$p = ρ R T$
Density is constant since V is constant and mass remains constant. R also has a constant. The pressure law can be used to simplify the problem.
$P 2 = P 1 (\frac{T 2}{T 1}) = 98.67$

eskalopit

Beginner2021-12-19Added 31 answers

p = ρ R T

(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 ψ 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} = (\frac{570 ° R}{520 ° R}) (90 ψ a) = 98.7 ψ a

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