- #1
barryj
- 853
- 51
Homework Statement
See attachment
I am having trouble getting my brain to understand what is happening here. See attachment.
Part a is easy since pv = nrt, Ta = pv/nr = (1E5)(.002)/(.05)(8.31) = 481K
It is not asked but likewise Tb = 2406 K, and Tc = 1203 K
Part b is the problem…. I need to fill in the blanks
I know,
for Pab there is no work since V is constant so heat added, Q, adds to U.
for Pbc, isothermal, there is no change in U, so Q must be equal to W
for Pca, W = p(Vc-Va) = (1E5)(.005 - .002) = 300 , problem states 150J of energy removed so Q = -150
I am confused by the signs of the variables in the equation deltaU = Q – W.
I assert the following.
Heat added results in a +Q while energy removed is a –Q. So Pab should have a positive Q and a +U.
Work done by the gas in say moving a piston should be +W. Since Pbc is isothermal, the Q must be equal to W resulting in no change in U.
Can someone straighten me out.