   # A compressed gas cylinder contains 1.00 × 10 3 g argon gas. The pressure inside the cylinder is 2050. psi (pounds per square inch) at a temperature of 18°C. How much gas remains in the cylinder if the pressure is decreased to 650. psi at a temperature of 26°C? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 62E
Textbook Problem
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## A compressed gas cylinder contains 1.00 × 103 g argon gas. The pressure inside the cylinder is 2050. psi (pounds per square inch) at a temperature of 18°C. How much gas remains in the cylinder if the pressure is decreased to 650. psi at a temperature of 26°C?

Interpretation Introduction

Interpretation: For the given data, mass of gas remains in the cylinder should be determined.

Concept introduction:

By combining the three gaseous laws namely Boyle’s law, Charles’s law and Avogadro’s law a combined gaseous equation is obtained. This combined gaseous equation is called Ideal gas law.

According to ideal gas law,

PV=nRT

Where,

P = pressure in atmospheres

V= volumes in liters

n = number of moles

R =universal gas constant ( 0.08206L×atm/K×mol )

T = temperature in kelvins

By knowing any three of these properties, the state of a gas can be simply identified with applying the ideal gas equation. For a gas at two conditions, the unknown variable can be determined by knowing the variables that change and remain constant and can be generated an equation for unknown variable from ideal gas equation.

### Explanation of Solution

Explanation

Ideal gas equation is modified at constant volume and the equation has given at two conditions. Formula for mass of gas is derived from the above equation.

According to ideal gas equation,

PV=nRT

By rearranging the above equation,

nTP=VR

R is a gas constant; here V is also a constant. For a gas at two conditions the equation can be written as:

n1T1P1=VR=n2T2P2orn1T1P1=n2T2P2 (1)

Since,Numberofmoles=massingramgrammolecularmassMassingram=numberofmoles(n)×grammolecularmassMultiplyingthenumeratoroflefthandsideandrighthandsideoftheequationwithgivenmolecularmass,massingramcanbecalculated.Sotheaboveequationbecomes;n1(molarmass)T1P1=n2(molarmass)T2P2mass1×T1P1=mass2×T2P2Fromtheaboveequation,massofgascanbecalculatedasfollows;Mass2=mass1×T1P2T2P1...(2)

To find out the final mass of gas , it is needed to take and write the given data and substitute their values in the equation(2). For two conditions problem, units for P and V just needed to be the same units and it is not needed to convert the standard units. But in the case of pressure, it must be converted to the Kelvin.

The given data and its values.

Mass1=1

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