ThermodynamicsHard
Question
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 PV 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 the List II and select the correct answer using the codes given blow the lists.
Match the paths in List I with the magnitudes of the work done in the List II and select the correct answer using the codes given blow the lists.
Options
A.P → 4
Q → 3
R → 2
S → 1
Q → 3
R → 2
S → 1
B.P → 4
Q → 3
R → 1
S → 2
Q → 3
R → 1
S → 2
C.P → 3
Q → 1
R → 2
S → 4
Q → 1
R → 2
S → 4
D.P → 1
Q → 3
R → 2
S → 4
Q → 3
R → 2
S → 4
Solution
(P) Process GE is isobaric
So work done = P |ᐃV| = P0 |(VG - VE)|
= P0 |(32V0 - V0)|
= 31V0V0
(Q) Process GH is isobaric so
So work done = P0 |(32V0 - 8V0)|
= 24 P0V0
(R) Since process FH is adiabatic so 32 P0V05/3 = P0VH5/3
& work done W
(S) Process FG is isothermal
so work done = nRT ln
= 160 P0V0 ln2.
So work done = P |ᐃV| = P0 |(VG - VE)|
= P0 |(32V0 - V0)|
= 31V0V0
(Q) Process GH is isobaric so
So work done = P0 |(32V0 - 8V0)|
= 24 P0V0
(R) Since process FH is adiabatic so 32 P0V05/3 = P0VH5/3
& work done W
(S) Process FG is isothermal
so work done = nRT ln
= 160 P0V0 ln2.
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