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

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