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

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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.



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