One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the p − V diagram. The processes involved are isochoric, isobaric, isothermal or adiab

Question

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

The processes involved are isochoric, isobaric, isothermal or adiabatic. Match the paths in List-I with magnitudes of the work done in List-II and select the correct answer using the codes given below the lists.

(ln2 = 0.70)

TestHub · Accepted Answer

Process : FG :

Isothermal expension

FH : Adiabatic expension

For process FG :

$p_{1} V_{1} = p_{2} V_{2}$

$32 p_{0} \times V_{0} = p_{0} \times V_{G}$

$V_{G} = 32 V_{0}$

For process FH :

$p_{1} V_{1}^{r} = P_{2} V_{2}^{r}$

$32 p_{0} \times V_{0}^{5/3} = p_{0} \times V_{H}^{5/3}$

$V_{H} = 8 V_{0}$

$∴ W_{GE} = - p_{0} [3 V_{0} - V_{0}] = 31 p_{0} V_{0}$

$∴ W_{GH} = - p_{0} [32 V_{0} - 8 V_{0}] = - 24 p_{0} V_{0}$

$∴ W_{FH} = \frac{p _{0} \times 8 V _{0} - 32 p _{0} V _{0}}{( \frac{5}{3} - 1 )} = - 36 p_{0} V_{0}$

$∴ W_{FG} = - 32 p_{0} V_{0} ln (\frac{32 V _{0}}{V _{0}}) = - 160 p_{0} V_{0} ln 2$

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