Suppose a 24.7 mL sample of helium gas at 25.°C and 1.14 atm is heated to 50°C and compressed to a volume of 14.8 mL. What will be the pressure of the sample?
My teacher got 2.06 atm. pls explain to me how he got this. THANK YOU
help me plsss. Explain
Help me pls
screw him, he's never gonna answer
Hey man, not funny, don't curse him dude
i'm not a dude and i didn't even curse. screw is not a curse word
sorry, i couldn't answer because i was busy. what was your question again?
(P1V1/T1) = P2V2/T2)
Don't forget T must be in kelvin.
Your teacher is correct at 2.06 atm.
tell me
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of temperature, volume, and pressure of a gas sample:
(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
Now, let's apply this equation to the given data:
P1 = 1.14 atm
V1 = 24.7 mL
T1 = 25°C = 298 K (temperature in Kelvin)
P2 = ?
V2 = 14.8 mL
T2 = 50°C = 323 K (temperature in Kelvin)
Plugging these values into the equation, we get:
(1.14 atm * 24.7 mL) / 298 K = (P2 * 14.8 mL) / 323 K
To solve for P2 (final pressure), we need to isolate it on one side of the equation. We can start by cross-multiplying and then solving for P2:
(1.14 atm * 24.7 mL * 323 K) = (P2 * 14.8 mL * 298 K)
(1.14 * 24.7 * 323) = (P2 * 14.8 * 298)
Now, divide both sides of the equation by (14.8 * 298) to solve for P2:
(1.14 * 24.7 * 323) / (14.8 * 298) = P2
Calculating this, we get:
P2 ≈ 2.06 atm
Therefore, the pressure of the sample at the new conditions (50°C and compressed to 14.8 mL) is approximately 2.06 atm.