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Problem 17 This figure (Figure 1) shows a container that is sealed at the top by

ID: 809444 • Letter: P

Question

Problem 17

This figure (Figure 1) shows a container that is sealed at the top by a movable piston. Inside the container is an ideal gas at 1.00 atm, 20.0 ?C, and 1.00 L. This information will apply to all parts of this problem A, B, and C.

Part A

What will the pressure inside the container become if the piston is moved to the 2.20 L mark while the temperature of the gas is kept constant?

Express your answer with the appropriate units.

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Part B

The gas sample has now returned to its original state of 1.00 atm, 20.0 ?C and 1.00 L. What will the pressure become if the temperature of the gas is raised to 200.0 ?C and the piston is not allowed to move?

Express your answer with the appropriate units.

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Part C

The gas described in parts A and B has a mass of 1.66 grams. The sample is most likely which monoatomic gas?

Type the elemental symbol of the gas below.

symbol =

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Figure 1 of 1

Problem 17

This figure (Figure 1) shows a container that is sealed at the top by a movable piston. Inside the container is an ideal gas at 1.00 atm, 20.0 ?C, and 1.00 L. This information will apply to all parts of this problem A, B, and C.

Part A

What will the pressure inside the container become if the piston is moved to the 2.20 L mark while the temperature of the gas is kept constant?

Express your answer with the appropriate units.

P =

SubmitHintsMy AnswersGive UpReview Part

Incorrect; One attempt remaining; Try Again

Part B

The gas sample has now returned to its original state of 1.00 atm, 20.0 ?C and 1.00 L. What will the pressure become if the temperature of the gas is raised to 200.0 ?C and the piston is not allowed to move?

Express your answer with the appropriate units.

P =

SubmitHintsMy AnswersGive UpReview Part

Part C

The gas described in parts A and B has a mass of 1.66 grams. The sample is most likely which monoatomic gas?

Type the elemental symbol of the gas below.

symbol =

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Figure 1 of 1

Explanation / Answer

Given data

Initial parameters of the ideal gas

Initial Volume of gas, V1 = 1.0 L

Initial pressure of gas P1 = 1.0 atm

Initial temperature of gas T1 = 20 0C

                                                    = (20 +273) K

                                                    = 293 K

A) In this case temperature T is kept constant and volume V is changed to 2.20 L

So Final volume V2 = 2.20 L

     Final pressure P2 = ?

According to Boyle's Law at constant temperature the pressure of the gas is inversely proportional to volume of the gas. This can be mathematically written as

       P = 1/V    or PV = constant

or P1V1 = P2V2

Therefore   P2 = (P1V1)/ V2

                     = (1.0 atm * 1.0L)/2.20 L

                     = 0.455 atm

Answer: pressure P = 0.455 atm

B) Here volume is returned to original but temperature is raised to 200 0C

        T2 =(200+273 = 473K)

     We know that for an ideal gas at constant volume

          (P1/T1) = (P2/T2)

                 P2 = (P1*T2) / T1

                      = (1.0 atm *473K) / 293 K

                      = 1.61 atm

Answer: pressure P = 1.61 atm

C) From Given data

Volume of gas, V = 1.0 L

Pressure of gas P = 1.0 atm

Tmperature of gas T = 20 0C

                              = (20 +273) K

                              = 293 K

mass of the gas = 1.66 g

We know that gas constant, R = 0.0821 L atm /mol K

We laso know that ideal gas equation is

          PV = nRT    where n = mole of the gas

            n = (PV) / (RT)

               = (1.0 atm* 1.0 L) / (0.0821 L atm/mol K * 293 K)

               = 0.0416 mol

Moles of the gas, n = mass of the gas / molar mass of the gas

          molar mass = mass of the gas/ moles of the gas

                            = 1.66g / 0.0416 mol

                            = 39.9 g/mol

This value is very close to molar mass of argon gas.

Hence the sample is most likely to be argon gas.

The elemental symbol of the gas = Ar