ample of helium gas has a volume of 3.10 L at 375. K and 82.0 atm. When the temperature changes to 210. K at 56.0 atm, what is the final volume? Be e your answer has the correct number of significant figures.
To solve this problem, we can use the combined gas law equation:
(P1V1)/T1 = (P2V2)/T2
Where:
P1 = initial pressure = 82.0 atm
V1 = initial volume = 3.10 L
T1 = initial temperature = 375 K
P2 = final pressure = 56.0 atm
V2 = final volume (unknown)
T2 = final temperature = 210 K
Let's plug in the values and solve for V2:
(82.0 atm * 3.10 L) / 375 K = (56.0 atm * V2) / 210 K
(254.2 atm*L) / 375 K = (56.0 atm * V2) / 210 K
Cross multiply:
375K * 56.0 atm * V2 = 254.2 atm * 210 K
21000 atm*V2 = 53322 atm*K
Now, divide by 21000 atm to solve for V2:
V2 = 53322 atm*K / 21000 atm
V2 = 2.53 L
Therefore, the final volume of the helium gas at 210 K and 56.0 atm is 2.53 L. Since the initial volume had 3 significant figures, the final volume should also be reported with 3 significant figures, making the final answer 2.53 L.