Question

Doubt in sign convention for adiabatic process..

Answer 1

if i increase the temperature from T2 to T1 adiabatically, then what will be my expression for work done?

Answer 2

\[-\frac{ nR }{ \gamma - 1 } (T2 - T1)\] right?

Answer 3

and why is this called "adiabatic compression" ? temperature is increasing.. so gas expands right?

Answer 4

@ParthKohli

Answer 5

@imqwerty @mayankdevnani

Answer 6

yes thats correct
in adiabetic compression the temp of gas increases

Answer 7

How?

Answer 8

dQ=dU+dW

1) heat supplied \(dQ=0 ~; Q=0\)

2)change in internal energy= \(\Delta U=nC_v\left[T_2-T_1\right]\)
=\(\large \frac{nR}{\gamma -1}\left[T_2-T_1\right]\)

3) work done by gas -> \(W=- \Delta U\)
=\(\large \color{red}+\frac{nR}{\gamma -1}\left[T_2-T_1\right]\)

Answer 9

why plus?

Answer 10

oh sorry
i made a mistake there

that will be -

Answer 11

ok..so mathematically it minus -so work is done on the gas -so its compression... But what is the physical meaning of saying temperature increases in an adiabatic compression?

Answer 12

And one more thing.. this is the 3rd stroke of carnot engine...So -W3 (while calculating the net work done..)is the abv. expression i wrote right? but the book says:
-W3 is the negative of the abv expression..

Answer 13

How is that?

Answer 14

well,
work done on gas = -ve
work done by gas==+ve

Answer 15

yes i know that thats why i have included the neagtive sign in my expression..

Answer 16

*negative

Answer 17

you can write that as
\(W=\large \frac{nR}{\gamma -1}\left[T_1 - T_2\right]\)

and
temperature is nothing but measure of energy in the system. and to increase the pressure we need to compress the gas and therefore we apply a force against the pressure of gas and and moving it through a distance
force moved through a distance implies expending energy.
and this energy goes into the gas.
now the gas has more energy so the tempreature will rise

Answer 18

if we are compressing , then work is done on the system means it is -ve
\[\large \bf \triangle U=-(-W)=+W\]
as internal energy is a function of TEMPERATURE and it is increasing,so TEMPERATURE will also be increasing

Answer 19

and i like the @imqwerty 's reasoning too

Answer 20

ok.. now i understood the adiabatic compression part..but what is the general expression for work done in an adiabatic process without sign coventions? ..

Answer 21

i'm confused that if we talk about chemistry,sign reverses ! then everything reverses too

Answer 22

yep!

Answer 23

i think this will work-

\(W=\Large \frac{nR}{\gamma -1} \left[T_1-T_2\right]\)

if its compression then temp will increase so W= -ve
if its expansion then temp will decrease so W=+ve

Answer 24

i know !

Answer 25

but in chemistry,sign reverses

Answer 26

i mean
work done on system=+ve
work done by system=-ve

Answer 27

but in chemistry i read is "1- gamma" in the denominator how? (before applying sign convention)

Answer 28

*it is

Answer 29

are you saying for both phy and chem its the abv. expression ? (except for the sign ofcourse)

Answer 30

please respond..

Answer 31

wait ! i'm thinking

Answer 32

ok!

Answer 33

i'm getting confused in my doubt

Answer 34

i think @imqwerty 's method is good !

Answer 35

we use sign convention for getting temp increased or decreased

Answer 36

without sign convention, imqwerty's method is good

Answer 37

we haven't done the thermodynamics part in chemistry yet
but i looked it up and found out this-

\(W= \frac{nR}{\gamma -1}\left[T_2-T_1\right]\)

this is causing confusion right?

Answer 38

yeah

Answer 39

for compression, temp decreased
for expansion,temp increased

Answer 40

is it true for chem ?

Answer 41

wait can you write in Ti and Tf... coz here in this Q t2 -> initial temp..

Answer 42

okay
chemistry formula- \(W= \frac{nR}{\gamma -1}\left[T_f-T_i\right]\)

physics formula- \(W= \frac{nR}{\gamma -1}\left[T_i-T_f\right]\)

Answer 43

yeah !

Answer 44

for compression, temp decreased
for expansion,temp increased

is it true for chem ?

Answer 45

No..

Answer 46

Thanks @Rushwr came up here

Answer 47

@Rushwr solve my doubt

Answer 48

wait.. both are same..

Answer 49

how ?

Answer 50

cos in chem i have been given gamma-1 in the denominator..

Answer 51

*1-gamma

Answer 52

are you sure ?

Answer 53

for chemistry-
\(q=\Delta U +(-w)\)
\(\Delta U=+w\) because q=0


but in physics we take it like this- \(\Delta U=-w\)

Answer 54

@mayankdevnani What is your doubt ?

Answer 55

for chem,
work done on system = +ve
work done by system=-ve

Answer 56

right ?

Answer 57

oh.. before that... is the abv expression correct or not?

Answer 58

(the one on the very top)

Answer 59

after applying sign coventions...

Answer 60

If work is done by the system then it's positive and if work is done on the system it is negative.

Answer 61

i think it is true for physics

Answer 62

This is derived from the first law thermodynamics and I really don't think that one same law will change for 2 different subjects

Answer 63

pls @mayankdevnani before thatanser my doubt..

Answer 64

*answer

Answer 65

i think its correct

Answer 66

then in the book why it says negative of this?

Answer 67

what did you consider T1 and T2 as ?

Answer 68

in physics book?
maybe they adjusted the - sign inside somewhere

Answer 69

T2-> initial tmp
T1 -> final temp
plug these in the exp. @imqwerty wrote and apply sign conv.. then you get my answer..

Answer 70

correct

Answer 71

so.. the book is wrong? coz it really important.. as its the essential part of the carnot cycle