A gas occupying 10 mL at standard conditions is heated to 10◦C while the pressure is reduced to 0.92114 atm. What is the new volume occupied by the gas?
Answer in units of mL
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure = standard pressure = 1 atm
V1 = initial volume = 10 mL
T1 = initial temperature in Kelvin = 273 K (standard conditions)
P2 = final pressure = 0.92114 atm
V2 = final volume (what we need to find)
T2 = final temperature in Kelvin = T1 + ∆T
Note: We need to convert the given temperatures from Celsius to Kelvin by adding 273.
Plugging in the values we have:
(1 atm * 10 mL) / (273 K) = (0.92114 atm * V2) / (273 K + 10 K)
Simplifying the equation:
10 / 273 = (0.92114 * V2) / 283
Cross-multiplying:
273 * 0.92114 * V2 = 10 * 283
Solving for V2:
V2 = (10 * 283) / (273 * 0.92114)
V2 ≈ 10.615 mL
Therefore, the new volume occupied by the gas is approximately 10.615 mL.
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
In this case:
P1 = the pressure at standard conditions (1 atm)
V1 = the initial volume (10 mL)
T1 = the initial temperature (standard temperature: 273.15 K)
P2 = the final pressure (0.92114 atm)
V2 = the final volume (unknown)
T2 = the final temperature (10°C → 283.15 K)
Let's rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Now we can substitute the known values into the equation:
V2 = (1 atm * 10 mL * 283.15 K) / (0.92114 atm * 273.15 K)
Calculating this expression:
V2 = (2831.5 mL* K) / (251.2831010)
V2 ≈ 11.28 mL
Therefore, the new volume occupied by the gas is approximately 11.28 mL.
(P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin.