Question

Uranium hexafluoride (UF6) sublimes from a solid to a gas and boils at a very low temperature of 56.5 oC. UF6 is used in the enrichment process of the production of fuel in nuclear reactions. At 56.5 oC and 1.00 atm, what is the density (in g/L) of UF6? = 13g/l

A 2.4 L sample of gas is placed in a container at 25oC and 2 atm of pressure. If the temperature is raised to 50oC and the volume is changed to 5.6 L, what is the new pressure? P=0.929

A gas is placed in a balloon with a volume of 3.0 L at 35oC and 800 torr. What would be the new volume for the gas if placed under STP? V=2.8L

Please explain them I have the solutions to each but Mess up in the process so please include each step thank you in advance!

Answer 1

1) Ideal gas law is PV = nRT where n = mass m/ molar mass M

R = universal gas constant = 0.0821 L.atm/K/mol

PV = (m/M) R T

P = (m/V) (RT/M)

m/V = density

Hence,

P = dRT/M

d = PM/RT

Given that P = 1 atm

               T = 56.5 oC = 56.5 + 273 K = 329.5 K

Molar mass of UF6, M = 352.02 g/mol       

Then,

d = PM/RT

d = 1 atm x 352.02 g/mol / 0.0821 L.atm/K/mol x 329.5 K

= 13.01 g/L

Therefore,

density (in g/L) of UF6 = 13.01 g/L

2)

Initial conditions:

Pressure P1 = 2 atm

Volume V1 = 2.4 L

Temperature T1 = 25 ^oC = 25 + 273 K = 298 K

Final conditions:

Pressure P2 = ?

Volume V2 = 5.6 L

Temperature T2 = 50oC = 50 + 273 K = 323 K

We knaw that P1V1/T1 = P2V2/T2

                       P2 = (P1V1/T1) (T2/V2)

                           = ( 2 atm x 2.4 L/ 298 K ) ( 323 K/ 5.6 L)

                          = 0.929 atm

                     P2 =0.929 atm

Therefore, new pressure = 0.929 atm

3)  

Initial conditions:

Pressure P1 = 800 torr

Volume V1 = 3 L

Temperature T1 = 35 ^oC = 35 + 273 K = 308 K

Final conditions: (STP)

Pressure P2 = 1 atm = 760 torr

Volume V2 = ? L

Temperature T2 = 0 ^oC = 0 + 273 K = 273 K

We knaw that P1V1/T1 = P2V2/T2

                       V2 = (P1V1/T1) (T2/P2)

                           = ( 800 mmHg x 3 L/ 308 K ) ( 273 K/ 760 mmHg)

                          = 2.8 L

                     V2 = 2.8 L

Therefore,

new volume for the gas = 2.8 L