Question

A sample of a certain gas has a volume of 1.25 L at -125°C and 5.0 ATM. The gas is compressed at 50.0 ATM to a volume of 325 mL. What is the final temperature of the helium gas in ⁰C?

Answer 1

To solve this problem, we can use the combined gas law, which states:

(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

Let's plug in the given values: P1 = 5.0 ATM V1 = 1.25 L T1 = -125°C (Convert to Kelvin: T1 = -125 + 273.15 = 148.15 K) P2 = 50.0 ATM V2 = 325 mL (Convert to liters: V2 = 325 / 1000 = 0.325 L) T2 = ? (what we want to find)

Now we can rearrange the formula to solve for T2:

T2 = (P2 * V2 * T1) / (P1 * V1)

Substituting the values:

T2 = (50.0 * 0.325 * 148.15) / (5.0 * 1.25)

T2 ≈ 238.34 K

Finally, let's convert the temperature back to °C:

T2 ≈ 238.34 - 273.15 ≈ -34.81°C

Therefore, the final temperature of the helium gas is approximately -34.81°C.