logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Starting with same initial conditions an ideal gas expands from volume $V_1$ to volume $V_2$ in three different paths. The work done by the gas is $W_1$ if process is purely isothermal, $W_2$ if process is purely isobaric and $W_3$, if process is purely adiabatic. Which is correct?

$\begin {array} {1 1} (A)\;W_2>W_1>W_3 & \quad (B)\;W_2>W_3>W_1 \\ (C)\; W_1>W_2>W_3 & \quad (D)\;W_1>W_3>W_2 \end {array}$

1 Answer

If it is isobaric, then $W = -P \Delta V$
If it is isothermal, then no heat exchange takes place, but work is done by the gas
If it is adiabatic, then no heat or energy exchange takes place, i.e, no work is done.
Ans : (A)
answered Mar 17, 2014 by thanvigandhi_1
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X