Question

Help answering parts b) c) and d) of this question on thermodynamics

Answer 1

just c) and d) now

Answer 2

did you use

\[\delta U = \frac{3}{2} R \delta T\] for change in energy?

Answer 3

for part a) yes

Answer 4

ok.. for part b.. u used

\[PV = RT\] to find V1 and V2 and then used
\[W = \int\limits_{V1}^{V2}PdV = P[V_2- V_1]\]

Answer 5

yeah

Answer 6

in part C.. do u understand what kind of process that is ?

Answer 7

yeah kept in equilibrium, think ive just got the correct answer now

Answer 8

kept in equilibrium? :O.. no its not.. work is done alright!

Answer 9

I used \[C_p - C_v = nR \] where \[C_v = 3/2nR\]

then used Cp to get Qp using \[C_p = \frac{ Q_p }{ \Delta T }\]

then used Q_p in the first law to get W done on the system of 27020J

Answer 10

yes but done very slowly so that it stays in equilibrium

Answer 11

ok.. i think.. that works too :P.. i thought of using \[W = 2.303 nRT \log(V_2/V_1)\]

Answer 12

besides part D u must have the answer :P

Answer 13

Oh yeah, that'd be the simpler way :P

Answer 14

New to thermodynamics, yeah, just trying to do the final question now, can't see a way

Answer 15

the volume is fixed. !! .. thats the hint :P

Answer 16

so no work is done? and V_f = V_i when you integrate

Answer 17

exactly.. its an iso volumetric process.. hence no work done!

Answer 18

just to give u more for ur money.. try drawing a PV diagram for this will ya?

Answer 19

If the pressure stays the same and the volume stays the same, then i don't see a way how to draw it

Answer 20

it would just be one point on the PV diagram

Answer 21

no i mean.. draw the PV diagram for the entire cycle..

Answer 22

Oh okay

Answer 23

then u ll understand what happened to the gas..

Answer 24

also try calculating and drawing, in which process how much heat was rejected or absorbed