I forgot in my previous question to say 1014 millibars. The question is I live at 1014 millibars for barometric pressure and temp. is 61 degrees Fahrenheit. I have an imaginary balloon that contains 22 grams of CO2-what is the volume in liters of the balloon using the gas constant. I know I use the ideal gas law but how do I change the barometric pressure from millibars to atm
This will get you to mm Hg, then divide mm by 760 mm to change to atm.
http://www.google.com/search?q=1+bar+to+mm+Hg&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a
To change the barometric pressure from millibars (mb) to atmospheres (atm), you need to use the conversion factor that 1 atm is equal to 1013.25 mb. Here's how you can convert the given barometric pressure of 1014 mb to atm:
Barometric pressure (atm) = Barometric pressure (mb) / 1013.25
Using this conversion, the barometric pressure of 1014 mb is approximately 1.000741 atm.
Now let's proceed with calculating the volume of the imaginary balloon using the ideal gas law.
The ideal gas law is expressed as:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, convert the temperature from Fahrenheit to Kelvin:
T (K) = (T (°F) - 32) * (5/9) + 273.15
Given temperature T = 61°F, plugging it into the formula:
T (K) = (61 - 32) * (5/9) + 273.15
T (K) ≈ 288.15 K
Now, rearrange the ideal gas law formula to solve for volume:
V = (n * R * T) / P
Given:
n = 22 g CO2 (molar mass of CO2 = 44.01 g/mol)
R = 0.0821 L·atm/mol·K
T = 288.15 K
P = 1.000741 atm (converted barometric pressure from 1014 mb)
Converting grams of CO2 to moles:
n (mol) = 22 g CO2 / 44.01 g/mol
n ≈ 0.5 mol
Plugging in the values:
V = (0.5 mol * 0.0821 L·atm/mol·K * 288.15 K) / 1.000741 atm
Calculating this equation will give you the volume (V) of the imaginary balloon in liters.