Heat and Thermodynamics 5 Question 18

19. One mole of a monatomic ideal gas is taken along two cyclic processes $E \rightarrow F \rightarrow G \rightarrow E$ and $E \rightarrow F \rightarrow H \rightarrow E$ as shown in the $p-V$ diagram.

(2013 Adv.)

The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

List I List II
P. $G \rightarrow E$ 1. $160 p _0 V _0 \ln 2$
Q. $G \rightarrow H$ 2. $36 p _0 V _0$
R. $F \rightarrow H$ 3. $24 p _0 V _0$
S. $F \rightarrow G$ 4. $31 p _0 V _0$

Codes

P $Q$ $R$ $S$ $P$ $Q$ $R$ $S$
(a) 4 3 2 1 (b) 4 3 1 2
(c) 3 1 2 4 (d) 1 3 2 4
Show Answer

Answer:

Correct Answer: 19. (a)

Solution:

In $F \rightarrow G$ work done in isothermal process is

$$ \begin{aligned} n R T \ln \left(\frac{V _f}{V _i}\right) & =32 p _0 V _0 \ln \left(\frac{32 V _0}{V _0}\right) \\ & =32 p _0 V _0 \ln 2^{5}=160 p _0 V _0 \ln 2 \end{aligned} $$

In $G \rightarrow E, \Delta W=p _0 \Delta V=p _0\left(31 V _0\right)=31 p _0 V _0$

In $G \rightarrow H$ work done is less than $31 p _0 V _0$ i.e. $24 p _0 V _0$

$\ln F \rightarrow H$ work done is $36 p _0 V _0$



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