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Chapter 22: Problem 5

A 55 L cylinder contains \(A r\) at 145 atm and \(26^{\circ}\) C. What minimum volume of air at STP must have been liquefied and distilled to produce this Ar? Air contains \(0.934 \%\) Ar, by volume.

Short Answer

Expert verified
Using the aforementioned steps, determine the required volume of air at STP that was needed. The final computation will provide the specific volume in liters.

Step by step solution

01

Convert the temperature to Kelvin

The given temperature is \(26^{\circ}\) C. Converting it to Kelvin, we have \(T = 26 + 273.15 = 299.15\) K.
02

Use the Ideal Gas Law

Using the Ideal Gas Law equation \(PV = nRT\), and knowing that the molecular weight of Argon (Ar) is approximately 40 g/mol, the number of moles of Argon can be calculated.
03

Calculate the moles of Argon

First, solve the ideal gas law, \(PV = nRT\), for \(n\), where \(n\) is the number of moles, \(P\) is the pressure, \(V\) is the volume, \(R\) is the ideal gas constant, and \(T\) is the temperature. So, \(n = PV/RT\). Substitute the known values: \(P = 145\) atm, \(V = 55\) L, \(R = 0.0821\) L atm/mol K (value of gas constant), and \(T = 299.15\) K. This yields \(n = 145 \times 55 / (0.0821 \times 299.15)\). Calculate the value of \(n\).
04

Convert the volume percent to volume fraction

The volume percent of Argon in air is given as 0.934%. Converting this to a volume fraction, we have \(0.934 / 100 = 0.00934\). This is the volume fraction of Ar in the air.
05

Calculate the volume of air at STP

Finally, to find the volume of air that was needed to produce this Argon, divide the number of Argon moles by the volume fraction of Argon in air. As STP conditions define 1 mole of any gas to occupy 22.4 liters, multiply the obtained value by 22.4 to convert moles into liters at STP. Calculate the result and that gives the required volume of air at STP.

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