Processing math: 100%
All
BE/B.Tech
MBA/PGDM
MBBS
ME/M.Tech
B.Sc
BA
B.Com
BCA
BBA/BMS
B.Sc (Nursing)

Ranking

College ranked based on real data

Indiatoday - 1740
Collegedunia - 1406
IIRF - 1684
Outlook - 1318
NIRF - 1301
Top Ranked Colleges in India ›

Find Colleges

Discover 19000+ colleges via preferences

Best MBA colleges in India
Best BTech colleges in India
Discover Top Colleges in India ›

Compare Colleges

Compare on the basis of rank, fees, etc.

IIT Madras vs IIT Delhi
IIT Madras vs IIT Bombay
Compare Colleges ›

Exams

Know more about your exams

B.Com
B.Sc
B.Sc (Nursing)
BA
BBA/BMS
BCA
BE/B.Tech
Check All Entrance Exams in India ›

College Predictor

Know your college admission chances

JEE Main
JEE Advanced
CAT
NEET
GATE
NMAT
MAT
XAT
Find Where you may get Admission ›

Course Finder

Discover top courses in Indian Colleges 2025

BE/B.Tech - 963
MBA/PGDM - 1159
ME/M.Tech - 1221
B.Sc - 1052
Get Top Courses in Indian Colleges ›

Your Gateway to Top Colleges & Exams

Discover thousands of questions, past papers, college details and all exam insights – in one place.

Popular Colleges

Top Exams

JEE MainJEE Main
JEE AdvJEE Advanced
NEET UGNEET UG
BITSATBITSAT
COMEDKCOMEDK
VITEEEVITEEE
WBJEEWBJEE

One mole of an ideal gas at 300 K in thermal contact with its...

One mole of an ideal gas at 300 K in thermal contact with its surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of the surroundings (∆Ssurr) in J K-1 is:
(1 L atm = 101.3 J)
5.763
1.013
-1.013
-5.763
Solution:
From 1st law of thermodynamics,
qsys=∆U-w=0-[-Pext.∆V]
=3.0 atm×(2.0 L-1.0 L)=3.0 L atm
∴∆Ssurr=(qrev)surrT=-qsysT
=-3.0×101.3 J300 K
=-1.013 J/K
Alternate solution:
∆Ssurr=qsurrT=-qsysT=WsysT
∵ For the isothermal process:
∆U=0     qsys=-Wsys
∆Ssurr.=-Pext(Vf-Vi)T
=-3(2-1)300×101.3
=-1.013  J/K