Answer:
1. By Pressure factor: if we double the pressure volume become half of its original
2. 2.14 L
3. 2.15 L
Explanation:
part 1
Data Given:
volume of container change
temperature of remain constant
The pressure doubles
Solution:
This problem can be explained by Boyle's Law that at constant temperature pressure and volume has an inverse relation with each other.
So the volume change due to change in Pressure.
P1V1 = P2V2
if we consider conditions at STP, as follows
initial volume V1 = 22.42 L
and
initial pressure P1 = 1 atm
if the pressure doubles then
final pressure P2 = 2 atm
Put values in Boyle's law equation
(1 atm) (22.42L) = (2 atm) (V2)
Rearrange the above equation to find V2
V2 = (1 atm) (22.42L) / 2 atm
V2 = 11.12 L
So it is clear from calculation if we double the pressure volume become half of its original. So its the pressure due to volume become effected and decrease by its increase.
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Part 2
Data Given:
Initial temperature T1= 250 K
final Temperature T2= 350 K
initial volume V1 = ?
final volume V2 = 3.0 L
Solution:
This problem will be solved by Charles' Law equation at constant pressure
V1 / T1 = V2 / T2 . . . . . . . . (1)
put values in above equation
V1 / 250 K = 3.0 L / 350 K
Rearrange the above equation to calculate V1
V1 = (3.0 L / 350 K) x 250 K
V1 = (0.0086 L . K) x 250 K
V1 = 2.14 L
So the initial volume = 2.14 L
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part 3
Data Given:
Initial temperature T1= 20 ºC
Convert Temperature from ºC to Kelvin
T = ºC + 273
T = 20 + 273 = 293 K
final Temperature T2= -5.00 ºC
Convert Temperature from ºC to Kelvin
T = ºC + 273
T = - 5.00 + 273 = 268 K
initial volume V1 = 2.35 L
final volume V2 = ?
Solution:
This problem will be solved by Charles' Law equation at constant pressure
V1 / T1 = V2 / T2 . . . . . . . . (1)
put values in above equation
2.35 L / 293 K = V2 / 268 K
Rearrange the above equation to calculate V1
V2 = (2.35 L / 293 K) x 268 K
V2 = (0.008 L . K) x 268 K
V2 = 2.15 L
So the volume at -5.00ºC = 2.15 L
