### HMWK PBLM 27

greenspun.com : LUSENET : UR General Chemistry : One Thread

Does anyone know how to calculate the values for step two of Pathway One? Neither V nor P are constant!

-- Anonymous, February 04, 2000

Hey! wow, Tara and I had the same problem...we couldn't do the problem because of that part. If anyone could provide an answer that would be really great. thanks!

-- Anonymous, February 04, 2000

For step two you can use the values they give of P and V to find the temperature for each set of data by using PV = nRT. These two values of Temperature allow you to subtract them and find the change in temperature for step two. Knowing change in Temp allows you to find delta H and delta E, but I still don't know how to get q or w.

-- Anonymous, February 05, 2000

How did you obtain deltaE or deltaH by using this method??? I know how to obtain for this part: -you know that w=-p*deltaV. well, in this case, the pressure is going to be the value 6 based on the assumption given in the problem. -next to obtain deltaV you must subtract 55 from 20 to give a value of -35. -to multiply -p and delta V gives a total of 210L*atm. -convert this over to Joules or kJoules. -your final value for work should equal 21.3kJ or 21273J

-- Anonymous, February 05, 2000

Why do we use p=6 atm rather than p= 6atm-3atm (the change in pressure from pathway 1 to pathway 2) in Pathway 2 of Problem #27?

-- Anonymous, February 06, 2000

To simplify typing delta = d Liz is right using the idael gas law:

dT = C*d(PV)/ nR d(PV) = ((P2*V2) - (P1*V1)) P2 = the second pressure

Then for pathway 1 step 2: dE = n*(Speciic heat)*dT

-- Anonymous, February 06, 2000

Sorry ignore what I just wrote in the previous answer I acidently summited before I was ready. The following is correct.

To simplify typing delta = d

Liz is right using the ideal gas law: dT = d(PV)/nR

also useful for this problem d(PV) = ((P2*V2) - (P1*V1)); where P2 = the second pressure, ect.

Then for pathway 1 step 2: dE = n*(Specific heat(Cv))*dT = n*(3R/2)*(d(PV)/nR) = (3/2)*d(PV)

dH = n*(Specific heat(Cp))*dT = n8(5R/2)*(d(PV/nR) = (5/2)*d(PV)

It says in the problem to assume external pressure is constant and equal to final pressure. Therefore, for this step external pressure = Pext = 6atm.

w = -Pext*dV

Then to find q:

dE = q + w q = dE - w

-- Anonymous, February 06, 2000