Question

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 2.20 L mark while the temperature of the gas is kept constant? Express your answer with the appropriate units.

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?

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

Answer 1

a) According tho the ideal gas equation, for a constant temperature process:

P * V = constant

Using the given initial conditions and pressure P for final state, we get:

1 * 1 = P * 2.20

Solving, P = 0.454 atm

b) For a constant volume process, P/T = constant. using the initial values of P and T in K, Pressure P, we get:

1 / (273 + 20) = P / (273 + 200)

Solving, we get: P = 1.614 atm

c) Using the ideal gas equation n = PV / RT to get the moles of gas

1 * 1 / (0.0821 * 293) = 0.0416 moles

Molar mass of gas = mass of the gas / number of moles = 1.66 / 0.0416 = 39.9 g / mol

This corresponds to the molar mass of Argon gas