A sample of fluorine gas occupies 810 milliliters at 270 K and 1.00 atm. What volume does the gas occupy when the pressure is doubled, and the temperature increases to 400 K

The answer to your question is Volume = 600 ml

Explanation:

Data

Volume 1 = 810 ml

Temperature 1 = 270°K

Pressure 1 = 1 atm

Volume 2 = ?

Pressure = 2 atm

Temperature 2 = 400°K

Process

1.- To solve this problem, use the Combine gas law.

               P1V1 / T1 = P2V2 / T2

- Solve for V2

                V2 = P1V1T2 / T1P2

2.- Substitution

                 V2 = (1)(810)(400) / (270)(2)

3.- Simplification

                 V2 = 324000 / 540

4.- Result

                 V2 = 600 ml

600mL

Explanation:

The following were obtained from the question:

V1 = 810mL

P1 = 1atm

T1 = 270K

V2 =?

P2 = 2atm(since the pressure is doubled)

T2 = 400K

The new volume can be obtained by doing the following :

P1V1/T1 = P2V2/T2

1 x 810/270 = 2 x V2 /400

Cross multiply to express in linear form

270 x 2 x V2 = 810 x 400

540 x V2 = 324000

Divide both side by 540

V2 = 324000/540

V2 = 600mL

The new volume of the gas is 600mL

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.