A 100.0-mL sample of air is collected at 21 degrees Celsius and 772 mm Hg. What is the volume of air at STP?
I know that I have to change deg. Celsius to K and mm Hg to atm.
I'm stuck on what formula I should use... Should I use PV=nRT?
Yes, you are on the right track! The ideal gas law equation, PV = nRT, is indeed the formula you should use to solve this problem. Here's how you can apply it:
Step 1: Convert the temperature from degrees Celsius to Kelvin.
To convert Celsius to Kelvin, you need to add 273.15 to the Celsius temperature. In this case, 21 degrees Celsius + 273.15 = 294.15 K.
Step 2: Convert the pressure from millimeters of mercury (mm Hg) to atmospheres (atm).
To convert mm Hg to atm, you divide the pressure in mm Hg by 760 (since 760 mm Hg is equivalent to 1 atm). In this case, 772 mm Hg / 760 = 1.0158 atm.
Step 3: Determine the number of moles (n).
Since the volume of air (V) and the temperature (T) are already given, you only need to find the number of moles to apply in the equation. To find the number of moles, you can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Substitute the given values:
n = (1.0158 atm) x (100.0 mL) / (0.0821 L∙atm/(mol∙K) x 294.15 K)
Note: The molar gas constant (R) has a value of 0.0821 L∙atm/(mol∙K).
Step 4: Calculate the volume of air at STP.
To find the volume of air at STP (standard temperature and pressure), you need to calculate the volume at the standard conditions of 1 atmosphere (atm) and 273.15 Kelvin (K). Use the same equation and rearrange it to solve for V:
V = nRT/P
Substitute the values:
V = (n) x (0.0821 L∙atm/(mol∙K)) x (273.15 K) / (1 atm)
Plug in the value of n from step 3:
V = [(1.0158 atm) x (100.0 mL) / (0.0821 L∙atm/(mol∙K) x 294.15 K)] x (0.0821 L∙atm/(mol∙K)) x (273.15 K) / (1 atm)
Simplify and calculate the volume:
V = 84.61 mL
Therefore, the volume of air at STP is approximately 84.61 mL.