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

Question

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.

A) What will the pressure inside the container become if the piston is moved to the 1.80 L mark while the temperature of the gas is kept constant?

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.

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

Explanation / Answer

Ans :

A)

USing the ideal gas law , pV = nRT

we get the relation as :

p1V1 = p2V2

p1 = 1.00 atm

V1 = 1.00 L

V2 = 1.80 L

Putting values :

1.00 x 1.00 = p2 x 1.80

p2 = 0.55 atm

The pressure will become 0.55 atm

B) Using the ideal gas law , pV = nRT

we get the relation as :

p1T2 = p2T1

T1 = 20 degrees C = 293.15 K

T2 = 200 degrees C = 473.15 K

p1 = 1.00 atm

putting values :

1.00 x 473.15 = p2 x 293.15

p2 = 1.61 atm

C)

Using the ideal gas law again :

1.00 x 1.00 = n x 0.0821 x 293.15

n = 0.0415 moles

Molar mass of gas = 1.66 / 0.0415

= 40 g/mol

This corresponds to the molar mass of Argon gas. So the sample is most likely Argon gas.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.