Given 300.0 mL of a gas at 17.0 degrees C. What is its volume at 10.0 degrees C?
Vf = .293 L
To find the volume of the gas at 10.0 degrees C, you can use the combined gas law, which relates the initial and final conditions of temperature, volume, and pressure of a gas.
The combined gas law formula is:
(Pi * Vi) / Ti = (Pf * Vf) / Tf
Where:
- Pi and Pf are the initial and final pressures of the gas, respectively.
- Vi and Vf are the initial and final volumes of the gas, respectively.
- Ti and Tf are the initial and final temperatures of the gas, respectively.
In this case, we have:
- Pi is not given, so we can assume it remains constant.
- Vi is given as 300.0 mL, which we need to convert to liters (1 L = 1000 mL).
- Ti is given as 17.0 degrees C, which we need to convert to Kelvin (T(K) = T(°C) + 273.15).
- Pf is not given, so we can assume it remains constant.
- Vf is 0.293 L (given).
- Tf is given as 10.0 degrees C, which we need to convert to Kelvin.
Now let's substitute the known values into the combined gas law equation:
(Pi * Vi) / Ti = (Pf * Vf) / Tf
Pi * Vi = (Pf * Vf * Ti) / Tf
Since we assume Pi and Pf to be constant, we can write the equation as:
Vi / Ti = Vf / Tf
Rearranging the equation to isolate Vf:
Vf = (Vi * Tf) / Ti
Let's substitute the known values:
Vi = 300.0 mL = 0.3 L (converted)
Ti = 17.0 degrees C + 273.15 = 290.15 K (converted)
Tf = 10.0 degrees C + 273.15 = 283.15 K (converted)
Vf = (0.3 L * 283.15 K) / 290.15 K
Now, calculate the value of Vf:
Vf = 293.0 mL
Therefore, the volume of the gas at 10.0 degrees C is approximately 293.0 mL.