an isolated quantity of an ideal gas at 297.2 K has a volume of 14.96 L at a pressure of 1.63 atm.
What is the volume of this gas sample when the absolute temperature is reduced to one third and the pressure is divided by five?
p1•V1/T1 = p2•V2/T2
V2 = p1•V1•T2/T1•p2
thank you!!!
To solve this problem, we can use the combined gas law equation, which states:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure (1.63 atm)
V1 = initial volume (14.96 L)
T1 = initial temperature (297.2 K)
P2 = final pressure (1.63 atm divided by 5)
V2 = final volume (we need to find this)
T2 = final temperature (297.2 K divided by 3)
First, let's calculate the final pressure (P2):
P2 = (1.63 atm) / 5 = 0.326 atm
Next, let's calculate the final temperature (T2):
T2 = 297.2 K / 3 = 99.0667 K
Now we can rearrange the equation to solve for V2:
(P1 * V1) / T1 = (P2 * V2) / T2
Plug in the values:
(1.63 atm * 14.96 L) / 297.2 K = (0.326 atm * V2) / 99.0667 K
Now, solve for V2:
V2 = [(1.63 atm * 14.96 L) / 297.2 K] * (99.0667 K / 0.326 atm)
V2 ≈ 10.26 L (rounded to two decimal places)
Therefore, the volume of the gas sample when the absolute temperature is reduced to one third and the pressure is divided by five is approximately 10.26 L.