A 1.05 g sample of an ideal gas occupies 1.045 L at 20.0°C and 0.975 atm. What will be its volume at -8.6°C and 0.5 atm?
I used PV=nRT for this. substituted P for 0.5 then n for 1.05, R .0821 then T for 264.4 K
After doing this I got .0219 but it is incorrect, what am I doing wrong?
N is not 1.05, that is the mass in grams.
do this:
PV=nRT
PV/T=nR for ideal gasses, nR is a constant.
At the second conditions,
PV/T=nR Hmmm. One is lead immediatlely to the combined gas law
P1V1/T1=P2V2/T2
solve for V2
To solve this problem using the ideal gas law equation, PV=nRT, you need to make sure all the units are consistent. Here's the correct way to approach it:
Step 1: Convert temperatures to Kelvin
20.0°C + 273.15 = 293.15 K
-8.6°C + 273.15 = 264.55 K
Step 2: Set up the equation using the initial conditions:
P1V1=nRT1
Step 3: Solve for n, the number of moles:
n = (P1V1) / (RT1)
n = (0.975 atm * 1.045 L) / (0.0821 L.atm/mol.K * 293.15 K)
Step 4: Use the ideal gas law equation to find the new volume:
P2V2 = nRT2
V2 = (nRT2) / P2
V2 = (n * 0.0821 L.atm/mol.K * 264.55 K) / 0.5 atm
By substituting the values into the equation, you should get the correct answer for the final volume.