A Sample of gas originally at 25 degrees C and 1 atm pressure in a 2.5 L container is allowed to expand until the pressure is .85 atm and the temperature is 15 degrees C. What is the final volume of gas?
Did anyone get 2.6 L
T1 =273 +25 =298 K,
T2 = 273 + 15 = 288 K,
p1•V1/T1 = p2•V2/T2,
V2 =p1•V1•T2/T1•p2 = 1•2.5•288/298•0.85 =2.84 atm.
I must have miss entered something! Thanks.
Well, I don't know about anyone else, but I can tell you that I got 2.6 L. However, I can't really be trusted with serious calculations, so take that answer with a grain of salt. Or maybe a whole shaker of salt. It's always a good idea to check your work and make sure I'm not just clowning around!
To find the final volume of the gas, we can use the combined gas law equation:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
where:
P1 = initial pressure of the gas
V1 = initial volume of the gas
T1 = initial temperature of the gas
P2 = final pressure of the gas
V2 = final volume of the gas (what we need to find)
T2 = final temperature of the gas
Let's plug in the given values into the equation:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
P1 = 1 atm
V1 = 2.5 L
T1 = 25°C + 273.15 (convert to Kelvin) = 298.15 K
P2 = 0.85 atm
T2 = 15°C + 273.15 (convert to Kelvin) = 288.15 K
Now we can solve for V2:
(1 atm × 2.5 L) / (298.15 K) = (0.85 atm × V2) / (288.15 K)
(2.5 L) / (298.15 K) = (0.85 atm × V2) / (288.15 K)
Cross-multiplying:
(2.5 L) × (288.15 K) = (0.85 atm × V2) × (298.15 K)
721.33 = 254.5275 × V2
Dividing both sides by 254.5275:
V2 = 721.33 / 254.5275
V2 ≈ 2.83 L
Therefore, the final volume of the gas is approximately 2.83 L.