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 do
ubled, and the temperature increases to 400 K
2 answers:
Answer:
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
Answer:
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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