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?
(P1V1/T1)=(P2V2/T2)
Remember T is in Kelvin.
To find the volume that the gas will occupy at a different pressure and temperature, we can use 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 expressed in Kelvin.
Let's begin by converting the temperatures from degrees Celsius to Kelvin. The Kelvin temperature scale is obtained by adding 273.15 to the Celsius value. So, in this case, we have:
T1 = 10 + 273.15 = 283.15 K (initial temperature)
T2 = 25 + 273.15 = 298.15 K (final temperature)
Now we can plug in the given values and solve for V2:
(200. mm Hg * 30. L) / (283.15 K) = (300. mm Hg * V2) / (298.15 K)
To simplify the calculation, we can cross-multiply:
200. mm Hg * 30. L * 298.15 K = 300. mm Hg * V2 * 283.15 K
Now, divide both sides of the equation by 300. mm Hg * 283.15 K to solve for V2:
V2 = (200. mm Hg * 30. L * 298.15 K) / (300. mm Hg * 283.15 K)
Canceling out the common units:
V2 ≈ (20 * 30) L ≈ 600 L
Therefore, the volume of the gas at 300. mm Hg and 25 degrees C is approximately 600 liters.