Question

Suppose a piston automatically adjusts to maintain a gas at a constant pressure of 12.20 atm . For the initial conditions, consider 0.05 mol of helium at a temperature of 230.00 K . This gas occupies a volume of 0.08 L under those conditions. What volume will the gas occupy if the number of moles is increased to 0.08 mol (n2) from the initial conditions? What volume will the gas occupy if the temperature is increased to 330.00 K (T2) from the initial conditions? Remember to reset the experiment to the initial conditions before determining each new volume. Express the volumes in liters to two decimal places separated by a comma.

Answer 1

Answer:

Situation 1: the new volume is 0.13 L

Situation 2: the new volume is 0.11 L

Explanation:

Step 1: Data given

The initial pressure = 12.20 atm

Number of moles of Helium = 0.05 moles

Temperature = 230.00 K

The volume 0.08 L

Situation 1: The number of moles increases to 0.08 moles

p*V = n*R*T

⇒with p = the initial pressure = 12.20 atm

⇒with V = the initial volume = 0.08 L

⇒with n = the initial number of moles of helium = 0.05 moles

⇒with R = the initial gas constant = 0.08206 L*atm/mol * K

⇒with T = the initial temperature = 230.00 K

In this situation the number of moles changes and we will calculate the volume

V1/n1 = V2/n2

⇒with V1 = the initial volume = 0.08 L

⇒with n1 = the initial number of moles = 0.05 moles

⇒with V2 = the new volume = TO BE DETERMINED

⇒with n2 = the increased number of moles = 0.08 moles

0.08 L / 0.05 moles = V2 / 0.08 moles

V2 = 0.13 L

What volume will the gas occupy if the temperature is increased to 330.00 K (T2) from the initial conditions?

V1 / T1 = V2/ T2

⇒with V1 the initial = The initial volume = 0.08 L

⇒with T1 = the initial temperature = 230.00 K

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the increased temperature = 330.00 K

0.08 L / 230.00 K = V2 / 330.00 K

V2 = 0.11 L