Discussion Forum

One mole of a monoatomic 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 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.

Answer:

Explanation:

in F→ G work done in isothermal process is 

$nRT ln\left(\frac{V_{f}}{V_{i}}\right)=32p_{0}V_{0}ln\left(\frac{32V_{0}}{V_{0}}\right)$

  $=32p_{0}V_{0}ln2^{5}=160p_{0}V_{0}ln 2$

$In G\rightarrow E, \triangle W=p_{0}\triangle V=p_{0}(31V_{0})=31p_{0}V_{0}$

 In G $\rightarrow$ H , work done is less than 31 p₀V₀ , i.e, 24 p₀V₀

 In F $\rightarrow$ H workdone is 36 p₀V₀