Helium gas, He, at 22 degrees C and 1.00 atm occupied a vessel whose volume was 2.54 L. What volume would this gas occupy if it were cooled to liquid-nitrogen temperature (-197 degrees celcius).
What is the equation?
Always start with PV/nT = PV/nT, where P is the pressure, V is the volume, n is the number of moles, and T is the temperature. Take out whichever terms you don't need. In this case, you don't need n (the number of moles), so you're left with:
PV/T = PV/T
Solve for the V on the right side of the equation after you plug in everything else.
( You only use PV=nRT when none of the variables change. In this case the temperature is changing, so you use PV/nT = PV/nT )
I did that and got -22.7.
1st of all the answer is supposed to be .65 L
2nd of all where does the negative go?
Temperature must always be in Kelvin! Add 273 to the temperature in Celsius to get the temperature in Kelvin. In this case:
INITIAL TEMPERATURE = 22+273 = 295 K
FINAL TEMPERATURE = -197+273 = 76 K
ooh ok that explains it lol thanks =]
To solve this problem, you can use the ideal gas law equation, which is given as:
PV = nRT
Where:
P is the pressure of the gas (in atmospheres)
V is the volume of the gas (in liters)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas (in Kelvin)
First, we need to convert the given temperatures to Kelvin:
22 degrees Celsius + 273 = 295 K (initial temperature)
-197 degrees Celsius + 273 = 76 K (final temperature)
Let's label the initial conditions as 1 and the final conditions as 2:
P1 = 1.00 atm
V1 = 2.54 L
T1 = 295 K
We want to find V2, the final volume.
Using the ideal gas law, we have:
P1V1/T1 = P2V2/T2
Now we need to rearrange the equation to solve for V2:
V2 = (P1V1 * T2) / (P2 * T1)
Plugging in the given values:
V2 = (1.00 atm * 2.54 L * 76 K) / (1.00 atm * 295 K)
Calculating the result:
V2 ≈ 0.685 L
Therefore, the volume of helium gas at liquid nitrogen temperature would be approximately 0.685 L.