Question

Please Help!!

1. When the volume of a container of gas changes by a certain factor at a constant temperature, the pressure doubles. By what factor does the container’s volume change?
2. The temperature of a gas rose from 250K to 350K. At 350K, the volume of the gas was 3.0L. If the pressure did not change, what was the initial volume of the gas?
3. A sample of a gas takes up 2.35L of space at room temperature (20.0ºC). What volume will the gas occupy at -5.00ºC? (Hint: Don’t forget to convert the temperatures to kelvins.)

Answer 1

1. By Pressure factor: if we double the pressure volume become half of its original

2. 2.14 L

3. 2.15 L

Step-by-step 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.

_____________

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

_________________

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