What will be the final volume of a 425 mL sample of gas initially at 22 degrees C and 1 atm pressure when the temperature is changed to 30 degrees C and the pressure to 0.471 atm?
PV = kT
V = kT/P
Given new P and T, call them P' and T', we have
V' = kT'/P'
V'/V = kT'/P' * P/kT = (T'/T)(P/P')
P' = .471P
T' = (30+273)/(22+273) = 1.0271T
So, we now must have
V'/V = 1.0271/.471 = 2.18
V' = 2.18 * 425 = 927 mL
To calculate the final volume of the gas sample, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (1 atm)
V1 = initial volume (425 mL)
T1 = initial temperature in Kelvin (22 degrees Celsius + 273.15 = 295.15 K)
P2 = final pressure (0.471 atm)
V2 = final volume (unknown)
T2 = final temperature in Kelvin (30 degrees Celsius + 273.15 = 303.15 K)
Plugging in the values:
(1 atm * 425 mL) / 295.15 K = (0.471 atm * V2) / 303.15 K
Now, let's solve for V2:
V2 = (1 atm * 425 mL * 303.15 K) / (295.15 K * 0.471 atm)
V2 = 1,273.87 mL
Therefore, the final volume of the gas sample will be approximately 1,273.87 mL.
To find the final volume of the gas sample, we can use the combined gas law equation, which relates the initial and final conditions of the gas sample.
The combined gas law equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure
V2 = Final volume (what we need to find)
T2 = Final temperature
Given:
P1 = 1 atm
V1 = 425 mL (or 0.425 L)
T1 = 22 degrees C (or 295 K)
P2 = 0.471 atm
T2 = 30 degrees C (or 303 K)
Let's substitute these values into the equation:
(1 atm * 0.425 L) / (295 K) = (0.471 atm * V2) / (303 K)
To find V2, we can rearrange the equation:
V2 = (1 atm * 0.425 L * 303 K) / (295 K * 0.471 atm)
Now, let's simplify the equation:
V2 = (128.275 atm L K) / (138.645 atm)
Finally, we can calculate the final volume:
V2 ≈ 0.9265 L
Therefore, the final volume of the gas sample will be approximately 0.9265 L.