KTGHard
Question
When a system is taken from state ′a′ to state ′b′ along the path ′acb′, it is found that a quantity of heat Q = 200 J is absorbed by the system and a work W = 80J is done by it. Along the path ′adb′, Q = 144J. The work done along the path ′adb′ is
Options
A.6J
B.12J
C.18J
D.24J
Solution
ᐃU = same is both process
Qacb − Wacb = Qadb − Wadb .
200 − 80 = 144 − Wadb. Wadb = 24 J.
Qacb − Wacb = Qadb − Wadb .
200 − 80 = 144 − Wadb. Wadb = 24 J.
Create a free account to view solution
View Solution FreeMore KTG Questions
A thermally insulated chamber of volume 2V0 is divided by a frictionless piston of area S & mass m into two equal parts ...The process ᐃU = 0, for an ideal gas can be best represented in the form of a graph :...The mean kinetic energy of a molecule of an ideal gas is :...The kinetic energy associated with per degree of freedom of a molecule is-...Gases O2 and H2 filled in a container the ratio at their r.m.s. velocity -...