Q. 4

# Consider the processes A and B shown in figure. It is possible that

A. both the processes are isothermal

B. both the processes are adiabatic

C. A is isothermal and B is adiabatic

D. A is adiabatic and B is isothermal

Answer :

From the graph, we can see that the slope of process B is steeper than that of process A.

Now in an isothermal process, under constant temperature, Pressure(P) x Volume(V) = constant … (i) (according to Boyle’s law)

Differentiating on both sides:

PdV + VdP = 0.

slope = = … (ii)

For an adiabatic process,

PVᵞ = constant

Where,

𝛾 is the ratio of specific heat of the gas at constant pressure and constant volume.

Differentiating the above question, we get

VᵞdP + 𝛾V^{𝛾}^{-1}PdV= 0

which is greater than that in case of the isothermal process.

Thus path (A) is for isothermal process while path (B) is for the adiabatic process.

Options (a) and (b) are incorrect since the slopes of the two paths Differ.

Option (d) is incorrect because the slope of path (A) is less than that of path (B).

Hence, the correct answer is option (c).

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