Ncert Exercises & Solutions

The figure shows the \((P\text-V)\) diagram of an ideal gas undergoing a change of state from \(A\) to \(B.\) Four different paths \(\mathrm{I, II, III}\) and \(\mathrm{IV},\) as shown in the figure, may lead to the same change of state.

(a)	The change in internal energy is the same in cases \(\mathrm{IV}\) and \(\mathrm{III}\) but not in cases \(\mathrm{I}\) and \(\mathrm{II}.\)
(b)	The change in internal energy is the same in all four cases.
(c)	The work done is maximum in case \(\mathrm{I}.\)
(d)	The work done is minimum in case \(\mathrm{II}.\)

Which of the following options contains only correct statements?

1.	(b), (c) and (d) only	2.	(a) and (d) only
3.	(b) and (c) only	4.	(a), (c) and (d) only

Hint: \(\Delta U = nC_V \Delta T\)

Step 1: Find the change in internal energy.
Given,
\(T_A =\frac{P_A V_A}{nR} ,T_B =\frac{P_B V_B}{nR}\)
In the given diagram all have the same initial and final state. Thus,
\(P_A V_A =P_B V_B \Rightarrow \frac{P_A V_A}{nR} =\frac{P_B V_B}{nR} \Rightarrow T_A=T_B\)

The change in internal energy for the process \(A\) to \(B,\) is given by;
\(\Rightarrow {dU}_{{A} \rightarrow {~B}}={nC}_{{V}} {dT}={nC}_{{V}}\left({~T}_{{B}}-{T}_{{A}}\right)\)
Since, \( T_A=T_B\)
\(\Rightarrow {dU}_{{A} \rightarrow {~B}}=0\)
Internal energy is a state function and does not depend on path, but on the initial and final state of the gas.
Thus change in internal energy is the same in all four cases.

Step 2: Find the work done.

The work done for \(A\) to \(B,\) is given by;
\(\Rightarrow {dW}_{{A} \rightarrow {~B}}= \int_A ^{B} {PdV}\)
\(\Rightarrow \int_A ^{B} {PdV}\) = Area under the \((P\text-V)\) curve from \(A\) to \(B.\)
It is evident that the area is maximum for the path \(I.\)
Thus, the work done is maximum in case \(I.\)
Hence, option (3) is the correct answer.

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