Answer to Question #242171 in Molecular Physics | Thermodynamics for Aizer

Question

A drum 0.5 ft in diameter and 45 inches long contained acetylene at 350 psia, 90 o F. After some of the

acetylene was used, the pressure was 200 psia and 85 o F. a) determine the amount of acetylene that

was used (lb m ). b) What volume would the used acetylene occupy at 14.7 psia, 80 o F (ft 3 ) ? R for this

gas is 59.35 ft-lb f /lb m- R.

Answer 1

The diameter of the drum is d = 0.5 ft

The length of the drum is l = 45 in × 1 12 ft 1 in = 3 . 75 ft

The initial pressure is P₁ = 350 psia × 144 lb / ft² 1 psia = 50400 lb / ft²

The initial temperature is T₁ = 90 ° F + 460 ° R = 550 ° R

The final pressure is P₂ = 200 psia × 144 lb / ft² 1 psia = 28800 lb / ft²

The final temperature is T₂ = 85 ° F + 460 ° R = 545 ° R

The gas constant is R = 59.35 ft-lb f /lb m -R

The volume of the drum is calculated as,

$V = \frac{π d^2 \times l}{4} = \frac{π \times 0.5^2 \times 3.75 }{4} \\ V = 0.7363\; ft^3$

(a) $P_1V =m_1RT_1$

The mass of acetylene initially in the drum is calculated as,

$m_1 = \frac{P_1 \times V}{ R \times T_1} = \frac{50400 \times 0.7363 }{ 59.35 \times 550} = 1.1368 \; lbm$

The mass of acetylene left in the drum is calculated as,

$m_2 = \frac{P_2 \times V }{ R \times T_2} = \frac{28800 \times 0.7363 }{ 59.35 \times 545} = 0.6555 \; lbm$

The mass of acetylene used is calculated as,

$m_3 = m_1 - m_2 = 1.1368 - 0.6555 = 0.4813 \;lbm$

(b) The pressure of acetylene is P₃ $= 14.7\; psia \times 144\; lb / ft^2 \;1\; psia = 2116.8 \; lb/ft^2$

The temperature of acetylene is T₂ = 80 °F + 460 °R = 540 °R

$P_3V=m_3RT_3$

The volume would the used acetylene occupy is calculated as,

$V = \frac{m_3 \times R \times T_3 }{ P_3} \\ = \frac{0.4813 \times 59.35 \times 540 }{ 2116.8} = 7.2870 \; ft^3$