I don't need to use a volume since the problem states that the volume stays the same. I can set the initial and final states equal to each other by re-arranging the equation to RT/V=P/n. Then the P/n for the initial and final state of the mixtures should equal each other.
The total pressure of a mixture of oxygen and hydrogen is 1 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.4 atm when measured at the same values of T and V as the original mixture. What was the composition of the original...
Physical Chemistry Problem!
The total pressure of a mixture of oxygen and hydrogen is 1 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.4 atm when measured at the same values of T and V as the original mixture. What was the...
Since the integral says "dV" doesn't that mean it only needs to be integrated with respect to the variable V while holding all other variables constant?
so far these are the answers I have come up with...
A. [-1/((x+y2)2)]dx - [2y/((x+y2)2)]dy
I started this problem by moving the denominator up but i am not sure that was the correct first move.
b. (2xy/x2y)dx + (x2/x2y)dy
c. I was not sure where to start with this problem.
total differentials!
[SIZE="4"]a.) f = 1/(x+y2)
[SIZE="4"]b.) f = ln(x2y)
[SIZE="4"]c.) P = [(RT)/(V-B)]-[A/(T1/2V2)]
what are the total differentials of these functions?
[b]1. The problem statement, all variables and given
integral of : (nRT)/(V-nb)dV from Vi to Vf\int ^{Vf}_{Vi} [SIZE="4"](nRT)/(V-nb)dV
I am really stuck on this one!
Do you need to move the denominator up first?
Homework Statement
d/dT (e-E0/(kT))
Homework Equations
d/dx(ex) = ex
The Attempt at a Solution
e-E0/(kT) * E0/kT2 ?
Do you use the chain rule on the exponent of e?