A gas is initially at 345 K, 821 torr, and occupies a volume of 5.85 L. What volume will the gas occupy if the temperature and pressure decrease, respectively, to 7 degrees celsius and 0.921 atm?
(P1V1/T1) = (P2V2/T2)
T must be in kelvin.
To solve this problem, we can apply the combined gas law equation:
(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
We are given the following information:
- P1 = 821 torr
- V1 = 5.85 L
- T1 = 345 K
- T2 = 7 degrees Celsius = 7 + 273.15 = 280.15 K
- P2 = 0.921 atm
Now, let's plug these values into the equation and solve for V2:
(821 torr * 5.85 L) / (345 K) = (0.921 atm * V2) / (280.15 K)
Note: We need to make sure that the pressures are in the same units, so let's convert 821 torr to atm by dividing it by 760 torr/1 atm:
(821 torr / 760 torr/1 atm * 5.85 L) / (345 K) = (0.921 atm * V2) / (280.15 K)
(1.07960526 atm * 5.85 L) / (345 K) = (0.921 atm * V2) / (280.15 K)
6.32786691 atm L / K = 0.921 * V2 / 280.15
Now, let's solve for V2:
V2 = (6.32786691 atm L / K * 280.15 K) / 0.921 atm
V2 = 1933.62 L
Therefore, the gas will occupy a volume of 1933.62 L when the temperature and pressure decrease to 7 degrees Celsius and 0.921 atm, respectively.