I'm absolutely terrible at these, ive never been able to understand them. Any help would be appreciated!
A sample of gas has a volume of 1.68 liters at 1atm pressure and 158k. What is its volume at STP?
A sample of gas occupies a colume of 0.683 liters at 25deg C. What folume would the gas occupy at 100deg C?
thanks ahead of time!
Help finding new volumes of gases
Moderators: Xen, expert, ChenBeier
use the gas laws... the second problem seems to just foloww the equation where pressure is constant since it is not mentioned. So basically, always use ideal gas equation which is PV=nRT where p is pressure v is volume n is no. of moles R is gas constant and T is temperature. So in first problem.
The sample of gas has V=1.68 P=1 T=158
So your equations PV =nRT becomes,
1atm * 1.68L = nR* 158K --> equation 1
For the second condition of STP,
P=1atm T=273.15 K (standard conditions of temeparature and pressure) V=??
Set up the equation:
1 atm * V L = nR*273 K ---> equation 2
Now, compare equation 1 and equation 2
since it is the same gas so the number of moles (n) is same in both cases and R is just a gas constant which is similar in both cases.. so if you divide both equations you get...
Eq1/Eq 2 = (1atm * 1.68L = nR* 158K) / (1 atm * V L = nR*273.15 K)
you get,
1.68L / V L = 158k/ 273.15k
so everything cancels out and you get
(1.68L * 273.15 K) / 158 K = V
V=2.9 L
(make sure the calculation i did it roughly)
for the second problem,,
consider pressure moles and R constant in PV=nRT
and basically compare
V1/T1 = V2/T2 where 1 refers to initial conditions at 25 and 2 towards 100 dgress. MAKE SURE YOU CONVERT THE CELSIUS TO KELVIN DEGREES WOULD NOT GIVE YOU CORRECT ANSWER
hope that helps..
-Sharan
The sample of gas has V=1.68 P=1 T=158
So your equations PV =nRT becomes,
1atm * 1.68L = nR* 158K --> equation 1
For the second condition of STP,
P=1atm T=273.15 K (standard conditions of temeparature and pressure) V=??
Set up the equation:
1 atm * V L = nR*273 K ---> equation 2
Now, compare equation 1 and equation 2
since it is the same gas so the number of moles (n) is same in both cases and R is just a gas constant which is similar in both cases.. so if you divide both equations you get...
Eq1/Eq 2 = (1atm * 1.68L = nR* 158K) / (1 atm * V L = nR*273.15 K)
you get,
1.68L / V L = 158k/ 273.15k
so everything cancels out and you get
(1.68L * 273.15 K) / 158 K = V
V=2.9 L
(make sure the calculation i did it roughly)
for the second problem,,
consider pressure moles and R constant in PV=nRT
and basically compare
V1/T1 = V2/T2 where 1 refers to initial conditions at 25 and 2 towards 100 dgress. MAKE SURE YOU CONVERT THE CELSIUS TO KELVIN DEGREES WOULD NOT GIVE YOU CORRECT ANSWER
hope that helps..
-Sharan