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Chapter 1: Problem 47

A closed tank having a volume of \(2 \mathrm{ft}^{3}\) is filled with \(0.30 \mathrm{lb}\) of a gas. A pressure gage attached to the tank reads 12 psi when the gas temperature is \(80^{\circ} \mathrm{F}\). There is some question as to whether the gas in the tank is oxygen or helium. Which do you think it is? Explain how you arrived at your answer.

Short Answer

Expert verified
The gas in the tank is most likely to be Oxygen, as the calculated constant R is closer to the specific gas constant of Oxygen.

Step by step solution

01

Convert units

Pressure given is in psi, it should be converted to \(lb/ft^2\). Similarly, temperature is given in Fahrenheit and it needs to be converted to Rankine. So, Pressure = \(12 psi * 144 = 1728 lb/ft^2\); and Temperature = \(80°F +460 = 540 R\). The conversion of Fahrenheit to Rankine involves adding 460.
02

Calculate the Gas Constant (R)

After converting to standard units, we're now ready to solve for the Gas Constant (R) using modified form of Ideal Gas law \(R = PV/mT\). Substituting the given values \(R = (1728 * 2) / (0.30 * 540) = 21.333 ft.lb/slug.°R\)
03

Compare R to known gas constants

Comparing the calculated R to the given known specific constants of Oxygen and Helium, it is determined that the calculated R is closer to the R of Oxygen (48.27 ft.lb/slug.°R) than that of Helium (93.13 ft.lb/slug.°R). Thus, it can be concluded that the gas in the tank is more likely to be Oxygen.

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Most popular questions from this chapter

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