Part A What will the pressure inside the container become if the piston is moved

Question

Part A

What will the pressure inside the container become if the piston is moved to the 2.00 Lmark while the temperature of the gas is kept constant?

Express your answer with the appropriate units.

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.

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.

Explanation / Answer

[The first part of the question is not mentioned,i am assuming it to be 'Inside the container is a ideal gas at 1.00 atm , 20.0 0C , and 1.00 L ]

Part A) P1V1 =P2V2

=> 1 atm x 1.00 L = P2 x 2.00 L

=> P2 =1/2 =0.5 atm.

Part B) T1=20+273 =293 K ,T2 =200+273 =473 K

P1 / T1 = P2 / T2
=>1 atm / 293 K = P2 / 473 K
=>P2 = 473/293 =1.61 atm

Part C )

Moles gas = PV / RT = 1.00 atm x 1.00 L / 0.08206 x 293 K =1/24.04358= 0.0416
Molar mass = 1.66 g / 0.0416 = 39.90 g/mol
That is, the sample contains monoatomic gas Ar (argon).

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