Alright, so we are going to use the ideal gas law to solve this one. That law simply states that
#PV=nRT#
In this equation, #P# is pressure in atm
#V# is volume in liters
#n# is the amount of gas in moles
#R# is a constant equal to #(0.08206L*atm)/(K*mol)#
And finally, #T# is temperature in kelvin
So we are trying to solve for #P#, so we need that on one side of the equation. We can do that by just dividing by #V#
#(PcancelV)/cancelV=(nRT)/V#
#P=(nRT)/V#
Ok, at this point lets start plugging variables in.
#V=10.0L#
#P=(nRT)/(10.0L)#
To convert #C^o to K# we just add 273
#K=C^o +273#
#K=200^oC+273#
#T=473K#
#P=(n*R*473K)/(10.0L)#
Now for the tricky part. We need to convert 32g of #C_2H_6# into moles. To do this we must first find the molar mass of #C_2H_6#, then convert 32g to moles.
The molar mass is going to equal #(30.07g)/(mol)# because we simply add the atomic mass of both elements:
#C=12.0107*2#
#H=1.00794 *6#
....................................
#(30.07g)/(mol)#
So now we need to find out how much moles 32g is
#n=(mol)/(30.07g)*32g#
#n=(mol)/(30.07cancelg)*32cancelg#
#n=(32mol)/(30.07)#
#n=1.1mol#
Ok, lets plug that in and the constant #(R)#. I'd like to note that I like to put the constant outside the equations as it makes it a lot easier.
#P=(n*R*473K)/(10.0L)#
#P=(1.1mol*473K)/(10.0L)(0.08206L*atm)/(K*mol)#
This is the fun part, were you cancel out all the conversion names. NOTE: since you are looking to #P#, we should have #atm# left over at the end
#P=(1.1cancel(mol)*473cancelK)/(10.0cancelL)(0.08206cancelL*atm)/(cancelK*cancel(mol))#
Now, finally, we just solve! Notice that atm is all that left....
#P=(1.1*473*0.08206*atm)/(10.0)#
#P=4.70atm#
Hope this helped! If you have any more questions feel free to ask them!
~Chandler Dowd
