Thermodynamics Question 255
Question: For an ideal gas graph is shown for three processes, Process 1, 2 and 3 are respectively.
Options:
A) Isobaric, adiabatic isochoric
B) Adiabatic, isobaric, isochoric
C) Isochoric, adiabatic, isobaric
D) Isochoric, isobaric, adiabatic
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Isochoric proceess $ dV=0 $
$ W=0 $ proceess 1 Isobaric: $ W=P\Delta V=nRAT $
Adiabatic $ |W|=\frac{nR\Delta T}{\gamma -1}0<\gamma -1<1 $ .
As work done in case of adiabatic process is more so process 3 is adiabatic and process 2 is isobaric.
BETA
