An experiment called for 4.83 L of sulfur dioxide at 0 degrees C and 1.00 atm. What would be the volume of this gas at 25 degrees C and 1.00 atm?
V1= 4.83 L
V2= ?
T1= 273 K
T2= 298 K
1 atm P1 and P2
V2 = V1 * T2/T1
4.83 L * 298 K /273 K =
ANSWER: 5.27 L
Is this correct.
yes
Well, let me check my calculations.
4.83 L * 298 K / 273 K = 5.27 L
Yep, that looks correct! So the volume of sulfur dioxide at 25 degrees C and 1.00 atm would be approximately 5.27 L. But don't worry, you don't need to memorize this formula for the next time you want to chill some sulfur dioxide.
Yes, that is correct. The volume of the gas at 25 degrees C and 1 atm would be 5.27 L.
Yes, your calculation is correct! To find the volume of the gas at a different temperature, you can use the formula V2 = V1 * T2/T1, where V1 is the initial volume, T1 is the initial temperature, T2 is the final temperature, and V2 is the final volume.
In this case, the initial volume (V1) is given as 4.83 L, the initial temperature (T1) is 0 degrees C which is equivalent to 273 K, the final temperature (T2) is 25 degrees C which is equivalent to 298 K, and the final pressure (P2) is 1.00 atm, which is the same as the initial pressure (P1).
By substituting these values into the formula, you get:
V2 = 4.83 L * (298 K / 273 K) = 5.27 L
Therefore, the volume of the gas at 25 degrees C and 1.00 atm would be 5.27 L.