Home » Ask & Discuss » Physics. » Thermal Physics « Back to Discussion

Thermal Physics

Hot goIITian

Joined:	26 Apr 2008
Post:	195

10 Feb 2009 16:42:22 IST

0 People liked this

551

IMP Q--PLZZZZZ HELP---PRB IN ISOTROPIC PROCESSES---GUIDE-RATES ASSURED

None

A MONOTONIC GAS EXPANDS ACCORDING TO PROCESS PT^2/5 =CONSTANT............

(A)TEMP. OF GAS INCREASES.

(B)TEMP. OF GAS DECREASES

(C)NO HEAT EXCHANGE WITH GAS

(D)HEAT IS ABSORBED BY GAS

HOW 2 SOLVE SUCH TYPE OF QSTN..........

WAT R THE STEPS TAKEN 2 SOLVE SUCH PRB.........

PLZ SM1 ANS .............

M WAITIN......

SAME Q CAME IN AITS 2 TIMES......MUST B IMP

Share this article on:

Comments (16)

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

10 Feb 2009 19:07:56 IST

Like 1 people liked this

PV = nRT

so T^2/5 = (PV/nR)^2/5

Given: PT^2/5= k (constant)

Upon substition, we get:

P^(7/5)V^(2/5) = constant
P⁷V² = constant
PV^2/7 = constant
This is a reversible adiabatic process, so the answer is (C) no heat exchange with gas

Vivek Sharma

Hot goIITian

Joined: 9 Jan 2009

Posts: 123

10 Feb 2009 19:11:04 IST

Like 0 people liked this

how are you so sure that it is adiabatic !!!!

Vivek Sharma

Hot goIITian

Joined: 9 Jan 2009

Posts: 123

10 Feb 2009 19:13:20 IST

Like 0 people liked this

Are all questions of type PVⁿare adiabatic type , please explain your answer ...

Vivek Sharma

Hot goIITian

Joined: 9 Jan 2009

Posts: 123

10 Feb 2009 19:17:43 IST

Like 0 people liked this

Please give answer quickly ...

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

10 Feb 2009 19:21:18 IST

Like 0 people liked this

P.V^gamma = constant is for adiabatic process.

Vivek Sharma

Hot goIITian

Joined: 9 Jan 2009

Posts: 123

12 Feb 2009 01:25:37 IST

Like 0 people liked this

This is only true if gamma corresponds to a ideal gas !!!!like 1.66 for monoatomic and 1.4 for diatomic

Rohit Khattar

Cool goIITian

Joined: 10 Jan 2009

Posts: 49

12 Feb 2009 01:34:13 IST

Like 0 people liked this

i wud lyke to corect vivek here...it is not necessary that gamma shud have specific values as u have given..if the given sample is a mixture of gases gamma can have any value which could be 2/7 as in the above process......and certainly solution given by gaurav is perfect...and btw for more clarification refer to HCV part 2 - thermodynamic processes...

Vivek Sharma

Hot goIITian

Joined: 9 Jan 2009

Posts: 123

12 Feb 2009 01:42:09 IST

Like 0 people liked this

SO DO YOU MEAN THAT FOR EVERY PROCESSES IN WHICH P,V,T ARE LINKED IN POWER TO SOME CONSTANT , THEN IT IS A ADIABATIC PROCESS ????

A G Nikhil

New kid on the Block

Joined: 15 Dec 2008

Posts: 19

12 Feb 2009 12:58:14 IST

Like 0 people liked this

no thats not wat he told means. for a mixture of gases, gamma maybe different than the values like 1.66 and 1.4

A G Nikhil

New kid on the Block

Joined: 15 Dec 2008

Posts: 19

12 Feb 2009 13:02:08 IST

Like 0 people liked this

n for the question :

PT^(2/5) = const

therefore, since PV = nRT,

T^(7/5)/V = constant this implies that if V increases T also increases hence A is correct

v know that,

del H = del U + del W = nR(del T) + P(del V)

since (del T) is positive and P(del V) can be found out to be +ve, heat is absorbed by the gas... D is also correct

Vivek Sharma

Hot goIITian

Joined: 9 Jan 2009

Posts: 123

12 Feb 2009 17:36:05 IST

Like 0 people liked this

so , is Mr Spidey wrong ????

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

13 Feb 2009 13:05:13 IST

Like 0 people liked this

Nope, not wrong.. the answer is both B and C.. C as i showed above.. and B as the guy above has shown... Temperature decrease doesn't mean that energy is lost ( in case of adiabatic processes). The temperature decrease is converted into volumetric changes.

panda

Scorching goIITian

Joined: 25 Jan 2009

Posts: 255

19 Feb 2009 15:41:51 IST

Like 0 people liked this

how is gamma = 2/7

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

19 Feb 2009 15:58:30 IST

Like 0 people liked this

gamma = Cp/Cv = 5/3 for monotonic/monoatomic IDEAL gas

(check wiki: http://en.wikipedia.org/wiki/Adiabatic_process )

So PV^2/7 = constant as i proved in my first post (so no heat exchange)... And from that we also get T.V^-5/7 = constant which means temp. of gas increases...

(answer is given as BC instead of BA though. Might be wrong)

panda

Scorching goIITian

Joined: 25 Jan 2009

Posts: 255

19 Feb 2009 16:04:48 IST

Like 0 people liked this

for adiabetic it must have been PV^5/3 = constant.

Madmax

Blazing goIITian

Joined: 24 Dec 2007

Posts: 1111

20 Feb 2009 19:40:30 IST

Like 1 people liked this

dude, for the final time.........GAMMA =5/3 ONLY FOR IDEAL MONOATOMIC GASES!!!!!!

READ THE QUESTION.......ITS NOT MENTIONED ANYWHERE THT GAS IS IDEAL!!!!!!

Continuation to Spidey's answer (Read tht before reading the next para) :

the answer is BC......see U have proved tht it is an adiabatic process....ie.no heat exchange with gas and surroundings........and tht means the energy for expansion is taken from the internal energy of the system....this means tht the temp must decrease....excuse me if I'm wrong, but I'd strongly go for BC as my answer!!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team