a 30. L sample of gas exerts 200. mm Hg pressure at 10 degrees C. What volume does the gas have at 300. mm Hg and 25 degrees C?
17 Liters
(P1V1/T1)=(P2V2/T2)
Don't forget T is in Kelvin.
To solve this problem, we can use the combined gas law equation, which is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
We are given the following information:
P1 = 200 mm Hg
V1 = 30 L
T1 = 10 degrees C = 283.15 K (convert to Kelvin)
P2 = 300 mm Hg
T2 = 25 degrees C = 298.15 K (convert to Kelvin)
Now we can plug in these values into the equation and solve for V2:
(200 mm Hg * 30 L) / (283.15 K) = (300 mm Hg * V2) / (298.15 K)
Let's solve for V2:
(200 mm Hg * 30 L * 298.15 K) = (300 mm Hg * V2 * 283.15 K)
(200 * 30 * 298.15) = (300 * V2 * 283.15)
1794900 = 84945V2
Divide both sides by 84945:
V2 = 1794900 / 84945
V2 ≈ 21.12 L
Therefore, the volume of the gas at 300 mm Hg and 25 degrees C is approximately 21.12 L.
To determine the volume of the gas at different conditions, you can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures of the gas, respectively.
V1 and V2 are the initial and final volumes of the gas, respectively.
T1 and T2 are the initial and final temperatures of the gas, respectively.
Let's plug in the given values:
P1 = 200. mm Hg
V1 = 30. L
T1 = 10 degrees C = 10 + 273 = 283 K
P2 = 300. mm Hg
T2 = 25 degrees C = 25 + 273 = 298 K
We need to find V2.
Now, rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Plug in the values:
V2 = (200. mm Hg * 30. L * 298 K) / (300. mm Hg * 283 K)
Simplify:
V2 = (59,600) / (84,900)
V2 ≈ 0.703 L
Therefore, the volume of the gas at 300. mm Hg and 25 degrees C is approximately 0.703 L.