A sample of gas occupies 29 L under a pressure of 1.3 atm. What would the resulting volume be if the pressure were increased to 3.9 atm if the temprature did not change?
P1V1 = P2V2
Post your work if you get stuck.
(1.3 atm)(29L)=(P2)(3.9)
9.66=P2
Correct; however, I would have rounded differently.
My answer waw 9.66666 and since 6 is over 5, I would add to the number to arrive at 9.7 L and not 9.66 L. You are allowed two significant digits (29, 1.3 and 3.9 all have two s.f. and in multiplication and division we are allowed the number with the least s.f. That means we may have two places). Thus I rounded to 9.7 Liters. Thanks for using Jiskha.
Thank you for your help.
To solve this question, we can use the combined gas law formula, which relates the initial and final conditions of a gas sample. The combined gas law formula is as follows:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 and P2 are the initial and final pressures, respectively,
V1 and V2 are the initial and final volumes, respectively,
T1 and T2 are the initial and final temperatures, respectively.
In this case, we are given:
P1 = 1.3 atm (initial pressure)
V1 = 29 L (initial volume)
P2 = 3.9 atm (final pressure)
T1 = T2 (temperature remains constant)
First, let's substitute the known values into the formula:
(1.3 atm * 29 L) / T1 = (3.9 atm * V2) / T1
Since the temperature remains constant, T1 and T2 will cancel out:
1.3 atm * 29 L = 3.9 atm * V2
Now, let's solve for V2 (final volume):
V2 = (1.3 atm * 29 L) / 3.9 atm
V2 ≈ 9.67 L
Therefore, the resulting volume would be approximately 9.67 L if the pressure were increased to 3.9 atm while keeping the temperature constant.