Please help me get started, i'm not asking for the answer, just some help!
A gas occupies 340mL at 273 K and 610 torr. What final temperature would be required to increase the pressure to 1 atm, the volume being held constant?
P1/T1 P2T2
610/273 = 760/T2
Solve for T2 (in kelvin).
Note that I changed 1 atm to 760 torr to agree with the 610 torr. Units must match.
Ok so i'm confused because T=273K is already in Kelvin.
Applicator lyapko use raylaguna.ru
Sure, I can definitely help you get started with this problem.
To find the final temperature required to increase the pressure from 610 torr to 1 atm, you would need to use the combined gas law. The combined gas law states that for a fixed amount of gas at constant volume, the ratio of the initial pressure to the initial temperature is equal to the ratio of the final pressure to the final temperature.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
In this case, the volume is being held constant, so we can simplify the equation to:
P1 / T1 = P2 / T2
To solve for the final temperature (T2), you can rearrange the equation:
T2 = (P2 * T1) / P1
Now, let's plug in the values given in the problem:
P1 = 610 torr
T1 = 273 K
P2 = 1 atm
Before we can proceed, we need to ensure that the pressure values are in the same unit. Since 1 atm = 760 torr, we need to convert the initial pressure from torr to atm:
P1 = 610 torr / 760 torr/atm = 0.8026 atm
Now we can substitute these values into the equation to find the final temperature:
T2 = (1 atm * 273 K) / 0.8026 atm
Finally, to arrive at the answer, divide 1 atm by 0.8026 atm and multiply it by 273 K.