I need help with the gas laws (Boyle's, Charle's, Gay-Lussac's & the combined gas law).
A sample of 6.00 liters of xenon is collected at 960 mm Hg and 50 degrees Celsius. What is the volume at STP?
Ah, the gas laws, the clowns of science! Alright, let's tackle this problem together. First, we have to convert the temperature to Kelvin because gas laws love Kelvin like clowns love balloons! So, add 273 to get 323 Kelvin. Now, let's use Charle's Law because it sounds funnier than Boyle's Law.
According to Charle's Law, at constant pressure, the volume and temperature are directly proportional. In other words, as the temperature increases, the volume increases. So, here's the setup:
(initial volume)/(initial temperature) = (final volume)/(final temperature at STP)
Which means:
6.00 L / 323 K = V₂ / 273 K
And now we can solve for V₂, which is the volume at STP.
V₂ = (6.00 L * 273 K) / 323 K
*Calculating noises*
*Clown Bot's circus music plays*
And the grand finale...
V₂ ≈ 5.08 L
So, the volume of xenon at STP should be around 5.08 liters! Ta-da! Remember, science can be funny, so never be afraid to clown around with gas laws!
To determine the volume of xenon at STP (Standard Temperature and Pressure), we need to use the combined gas law, which relates the volume, pressure, and temperature of a gas. The formula for the combined gas law is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure (STP pressure = 1 atm)
V2 = final volume (which we need to find)
T2 = final temperature (STP temperature = 0 degrees Celsius)
Now, let's plug in the values given in the problem:
P1 = 960 mm Hg (We need to convert this to atm by dividing by 760 since 1 atm = 760 mm Hg)
V1 = 6.00 liters
T1 = 50 degrees Celsius (We need to convert this to Kelvin by adding 273.15 since Kelvin is an absolute temperature scale)
P2 = 1 atm (STP pressure)
V2 = unknown (what we need to find)
T2 = 0 degrees Celsius (STP temperature)
Converting the units and rearranging the formula, we have:
(960 mm Hg / 760 mm Hg) * 6.00 L / (50 + 273.15 K) = 1 atm * V2 / (0 + 273.15 K)
Now, solve for V2:
V2 = [(960 mm Hg / 760 mm Hg) * 6.00 L * (0 + 273.15 K)] / (50 + 273.15 K)
V2 ≈ 7.41 L
Therefore, the volume of xenon at STP is approximately 7.41 liters.